{"id":2817,"date":"2024-06-10T17:15:03","date_gmt":"2024-06-10T17:15:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2817"},"modified":"2025-02-20T18:51:52","modified_gmt":"2025-02-20T18:51:52","slug":"defining-the-derivative-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/defining-the-derivative-fresh-take\/","title":{"raw":"Defining the Derivative: Fresh Take","rendered":"Defining the Derivative: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Calculate the slope of a tangent line to a curve and find its equation<\/li>\r\n\t<li>Find the derivative of a function at a given point<\/li>\r\n\t<li>Explain how velocity measures speed over time, and compare average velocity over a period with the exact speed at a specific moment<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Tangent Lines<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Secant Lines and Difference Quotients:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Slope of a secant line: [latex]m_{\\text{sec}} = \\frac{f(x) - f(a)}{x - a}[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Difference quotient with increment [latex]h[\/latex]: [latex]Q = \\frac{f(a+h) - f(a)}{h}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Tangent Lines:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Limit of secant lines as they approach a point<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Slope of tangent line: [latex]m_{\\text{tan}} = \\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a}[\/latex] or [latex]m_{\\text{tan}} = \\lim_{h \\to 0} \\frac{f(a+h) - f(a)}{h}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Local Linearity:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Near the point of tangency, the function appears linear<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Tangent line provides a good approximation of the function<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Equation of Tangent Line:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Use point-slope form: [latex]y - y_1 = m(x - x_1)[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex](x_1, y_1)[\/latex] is the point of tangency, [latex]m[\/latex] is the slope of the tangent line<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169738935464\">Find the equation of the line tangent to the graph of [latex]f(x)=\\dfrac{1}{x}[\/latex] at [latex]x=2[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739001198\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739001198\"]<\/p>\r\n<p id=\"fs-id1169739001198\">We can use the first definition from before, but as we have seen, the results are the same if we use the other definition.<\/p>\r\n<div id=\"fs-id1169738961345\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}m_{\\tan} &amp; =\\underset{x\\to 2}{\\lim}\\frac{f(x)-f(2)}{x-2} &amp; &amp; &amp; \\text{Apply the definition.} \\\\ &amp; =\\underset{x\\to 2}{\\lim}\\frac{\\frac{1}{x}-\\frac{1}{2}}{x-2} &amp; &amp; &amp; \\text{Substitute} \\, f(x)=\\frac{1}{x} \\, \\text{and} \\, f(2)=\\frac{1}{2}. \\\\ &amp; =\\underset{x\\to 2}{\\lim}\\frac{\\frac{1}{x}-\\frac{1}{2}}{x-2} \\cdot \\frac{2x}{2x} &amp; &amp; &amp; \\begin{array}{l}\\text{Multiply numerator and denominator by} \\, 2x \\, \\text{to} \\\\ \\text{simplify fractions.} \\end{array} \\\\ &amp; =\\underset{x\\to 2}{\\lim}\\frac{(2-x)}{(x-2)(2x)} &amp; &amp; &amp; \\text{Simplify.} \\\\ &amp; =\\underset{x\\to 2}{\\lim}\\frac{-1}{2x} &amp; &amp; &amp; \\text{Simplify using} \\, \\frac{2-x}{x-2}=-1, \\, \\text{for} \\, x\\ne 2. \\\\ &amp; =-\\frac{1}{4} &amp; &amp; &amp; \\text{Evaluate the limit.} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169739270491\">We now know that the slope of the tangent line is [latex]-\\frac{1}{4}[\/latex]. To find the equation of the tangent line, we also need a point on the line. We know that [latex]f(2)=\\frac{1}{2}[\/latex]. Since the tangent line passes through the point [latex](2,\\frac{1}{2})[\/latex] we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation [latex]y=-\\frac{1}{4}x+1[\/latex]. The graphs of [latex]f(x)=\\frac{1}{x}[\/latex] and [latex]y=-\\frac{1}{4}x+1[\/latex] are shown in Figure 6.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205200\/CNX_Calc_Figure_03_01_006.jpg\" alt=\"This figure consists of the graphs of f(x) = 1\/x and y = -x\/4 + 1. The part of the graph f(x) = 1\/x in the first quadrant appears to touch the other function\u2019s graph at x = 2.\" width=\"487\" height=\"321\" \/> Figure 6. The line is tangent to [latex]f(x)[\/latex] at [latex]x=2[\/latex][\/caption]\r\n\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739231715\">Find the slope of the line tangent to the graph of [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=4[\/latex].<\/p>\r\n<p>[reveal-answer q=\"3776221\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"3776221\"]<\/p>\r\n<p id=\"fs-id1169738973750\">Use either definition. Multiply the numerator and the denominator by a conjugate.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739236594\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739236594\"]<\/p>\r\n<p id=\"fs-id1169739236594\">[latex]\\dfrac{1}{4}[\/latex]<\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to this example. <br \/>\r\n<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=738&amp;end=892&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><br \/>\r\n<br \/>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.1DefiningTheDerivative738to892_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Defining the Derivative\" here (opens in new window)<\/a>.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">Find the equation of the line tangent to [latex]f(x) = x^2[\/latex] at [latex]x = 3[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\"><br \/>\r\n[reveal-answer q=\"948182\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"948182\"]<\/p>\r\n<p>Calculate the slope:<\/p>\r\n<p style=\"text-align: center;\">[latex]<br \/>\r\n\\begin{array}{rcl}<br \/>\r\nm_{\\text{tan}} &amp;=&amp; \\lim_{x \\to 3} \\frac{f(x) - f(3)}{x - 3} \\\\<br \/>\r\n&amp;=&amp; \\lim_{x \\to 3} \\frac{x^2 - 9}{x - 3} \\\\<br \/>\r\n&amp;=&amp; \\lim_{x \\to 3} (x + 3) \\\\<br \/>\r\n&amp;=&amp; 6<br \/>\r\n\\end{array}<br \/>\r\n[\/latex]<\/p>\r\n<p>Identify the point:<\/p>\r\n<p style=\"text-align: center;\">[latex](3, f(3)) = (3, 9)[\/latex]<\/p>\r\n<p>Use point-slope form:<\/p>\r\n<p style=\"text-align: center;\">[latex]y - 9 = 6(x - 3)[\/latex]<\/p>\r\n<p>Simplify to slope-intercept form:<\/p>\r\n<p style=\"text-align: center;\">[latex]y = 6x - 9[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2>The Derivative of a Function at a Point<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Definition of the Derivative:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">The derivative is the instantaneous rate of change of a function at a point<\/li>\r\n\t<li class=\"whitespace-normal break-words\">It's equivalent to the slope of the tangent line at that point<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Notations for the Derivative:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">[latex]f'(a)[\/latex] denotes the derivative of [latex]f(x)[\/latex] at [latex]x = a[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Two Equivalent Definitions:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">[latex]f'(a) = \\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a}[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]f'(a) = \\lim_{h \\to 0} \\frac{f(a+h) - f(a)}{h}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li>Formal definition of a derivative:<br \/>\r\n<p class=\"whitespace-pre-wrap break-words\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0The derivative of a function [latex]f(x)[\/latex] at a point [latex]a[\/latex] is defined as:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]f'(a) = \\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0or equivalently:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]f'(a) = \\lim_{h \\to 0} \\frac{f(a+h) - f(a)}{h}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0provided these limits exist.<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739001692\">For [latex]f(x)=3x^2-4x+1[\/latex], find [latex]f^{\\prime}(2)[\/latex] by using the first definition.<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739104712\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739104712\"]<\/p>\r\n<p id=\"fs-id1169739104712\">Substitute the given function and value directly into the equation.<\/p>\r\n<div id=\"fs-id1169739305050\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}f^{\\prime}(x)&amp; =\\underset{x\\to 2}{\\lim}\\frac{f(x)-f(2)}{x-2} &amp; &amp; &amp; \\text{Apply the definition.} \\\\ &amp; =\\underset{x\\to 2}{\\lim}\\frac{(3x^2-4x+1)-5}{x-2} &amp; &amp; &amp; \\text{Substitute} \\, f(x)=3x^2-4x+1 \\, \\text{and} \\, f(2)=5. \\\\ &amp; =\\underset{x\\to 2}{\\lim}\\frac{(x-2)(3x+2)}{x-2} &amp; &amp; &amp; \\text{Simplify and factor the numerator.} \\\\ &amp; =\\underset{x\\to 2}{\\lim}(3x+2) &amp; &amp; &amp; \\text{Cancel the common factor.} \\\\ &amp; =8 &amp; &amp; &amp; \\text{Evaluate the limit.} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739269938\">For [latex]f(x)=x^2+3x+2[\/latex], find [latex]f^{\\prime}(1)[\/latex].<\/p>\r\n<p>[reveal-answer q=\"708365\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"708365\"]<\/p>\r\n<p id=\"fs-id1169736618844\">Use either the first definition, the second, or try both.\u00a0<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739293550\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739293550\"]<\/p>\r\n<p id=\"fs-id1169739293550\">[latex]f^{\\prime}(1)=5[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=1434&amp;end=1537&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.1DefiningTheDerivative1434to1537_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Defining the Derivative\" here (opens in new window)<\/a>.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">Find [latex]f'(2)[\/latex] for [latex]f(x) = 3x^2 - 4x + 1[\/latex] using the second definition.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\"><br \/>\r\n[reveal-answer q=\"723055\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"723055\"]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex] \\begin{array}{rcl} f'(2) &amp;=&amp; \\lim_{h \\to 0} \\frac{f(2+h) - f(2)}{h} \\\\ &amp;=&amp; \\lim_{h \\to 0} \\frac{[3(2+h)^2 - 4(2+h) + 1] - [3(2)^2 - 4(2) + 1]}{h} \\\\ &amp;=&amp; \\lim_{h \\to 0} \\frac{[12 + 12h + 3h^2 - 8 - 4h + 1] - [12 - 8 + 1]}{h} \\\\ &amp;=&amp; \\lim_{h \\to 0} \\frac{3h^2 + 8h}{h} \\\\ &amp;=&amp; \\lim_{h \\to 0} (3h + 8) \\\\ &amp;=&amp; 8 \\end{array} [\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore, [latex]f'(2) = 8[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2>Velocities and Rates of Change<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Average Velocity:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">For a position function [latex]s(t)[\/latex], average velocity over [latex][a,t][\/latex] is: [latex]v_{\\text{avg}} = \\frac{s(t) - s(a)}{t - a}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Instantaneous Velocity:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Limit of average velocity as time interval approaches zero<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Equivalent to the derivative of the position function: [latex]v(a) = s'(a) = \\lim_{t \\to a} \\frac{s(t) - s(a)}{t - a}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Instantaneous Rate of Change:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Generalization of instantaneous velocity to any function<\/li>\r\n\t<li class=\"whitespace-normal break-words\">For a function [latex]f(x)[\/latex], the instantaneous rate of change at [latex]a[\/latex] is [latex]f'(a)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Graphical Interpretation:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Average velocity: Slope of secant line<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Instantaneous velocity: Slope of tangent line<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739220848\">A rock is dropped from a height of [latex]64[\/latex] feet. Its height above ground at time [latex]t[\/latex] seconds later is given by [latex]s(t)=-16t^2+64, \\, 0\\le t\\le 2[\/latex]. Find its instantaneous velocity [latex]1[\/latex] second after it is dropped, using the definition of a derivative.<\/p>\r\n<p>[reveal-answer q=\"403817\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"403817\"]<\/p>\r\n<p id=\"fs-id1169739189236\">[latex]v(t)=s^{\\prime}(t)[\/latex].\u00a0<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739298454\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739298454\"]<\/p>\r\n<p id=\"fs-id1169739298454\">[latex]-32[\/latex] ft\/sec<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=1660&amp;end=1763&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.1DefiningTheDerivative1660to1763_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Defining the Derivative\" here (opens in new window)<\/a>.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739253553\">A toy company can sell [latex]x[\/latex] electronic gaming systems at a price of [latex]p=-0.01x+400[\/latex] dollars per gaming system. The cost of manufacturing [latex]x[\/latex] systems is given by [latex]C(x)=100x+10,000[\/latex] dollars. Find the rate of change of profit when [latex]10,000[\/latex] games are produced. Should the toy company increase or decrease production?<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739274608\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739274608\"]<\/p>\r\n<p id=\"fs-id1169739274608\">The profit [latex]P(x)[\/latex] earned by producing [latex]x[\/latex] gaming systems is [latex]R(x)-C(x)[\/latex], where [latex]R(x)[\/latex] is the revenue obtained from the sale of [latex]x[\/latex] games. Since the company can sell [latex]x[\/latex] games at [latex]p=-0.01x+400[\/latex] per game,<\/p>\r\n<div id=\"fs-id1169739286461\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp=x(-0.01x+400)=-0.01x^2+400x[\/latex].<\/div>\r\n<p id=\"fs-id1169739302924\">Consequently,<\/p>\r\n<div id=\"fs-id1169739302928\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(x)=-0.01x^2+300x-10,000[\/latex].<\/div>\r\n<p id=\"fs-id1169739190488\">Therefore, evaluating the rate of change of profit gives<\/p>\r\n<div id=\"fs-id1169739190491\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}P^{\\prime}(10000)&amp; =\\underset{x\\to 10000}{\\lim}\\frac{P(x)-P(10000)}{x-10000} \\\\ &amp; =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-10000-1990000}{x-10000} \\\\ &amp; =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-2000000}{x-10000} \\\\ &amp; =100 \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169736587942\">Since the rate of change of profit [latex]P^{\\prime}(10,000)&gt;0[\/latex] and [latex]P(10,000)&gt;0[\/latex], the company should increase production.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate the slope of a tangent line to a curve and find its equation<\/li>\n<li>Find the derivative of a function at a given point<\/li>\n<li>Explain how velocity measures speed over time, and compare average velocity over a period with the exact speed at a specific moment<\/li>\n<\/ul>\n<\/section>\n<h2>Tangent Lines<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Secant Lines and Difference Quotients:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Slope of a secant line: [latex]m_{\\text{sec}} = \\frac{f(x) - f(a)}{x - a}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Difference quotient with increment [latex]h[\/latex]: [latex]Q = \\frac{f(a+h) - f(a)}{h}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Tangent Lines:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Limit of secant lines as they approach a point<\/li>\n<li class=\"whitespace-normal break-words\">Slope of tangent line: [latex]m_{\\text{tan}} = \\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a}[\/latex] or [latex]m_{\\text{tan}} = \\lim_{h \\to 0} \\frac{f(a+h) - f(a)}{h}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Local Linearity:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Near the point of tangency, the function appears linear<\/li>\n<li class=\"whitespace-normal break-words\">Tangent line provides a good approximation of the function<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Equation of Tangent Line:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Use point-slope form: [latex]y - y_1 = m(x - x_1)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex](x_1, y_1)[\/latex] is the point of tangency, [latex]m[\/latex] is the slope of the tangent line<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738935464\">Find the equation of the line tangent to the graph of [latex]f(x)=\\dfrac{1}{x}[\/latex] at [latex]x=2[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739001198\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739001198\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739001198\">We can use the first definition from before, but as we have seen, the results are the same if we use the other definition.<\/p>\n<div id=\"fs-id1169738961345\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}m_{\\tan} & =\\underset{x\\to 2}{\\lim}\\frac{f(x)-f(2)}{x-2} & & & \\text{Apply the definition.} \\\\ & =\\underset{x\\to 2}{\\lim}\\frac{\\frac{1}{x}-\\frac{1}{2}}{x-2} & & & \\text{Substitute} \\, f(x)=\\frac{1}{x} \\, \\text{and} \\, f(2)=\\frac{1}{2}. \\\\ & =\\underset{x\\to 2}{\\lim}\\frac{\\frac{1}{x}-\\frac{1}{2}}{x-2} \\cdot \\frac{2x}{2x} & & & \\begin{array}{l}\\text{Multiply numerator and denominator by} \\, 2x \\, \\text{to} \\\\ \\text{simplify fractions.} \\end{array} \\\\ & =\\underset{x\\to 2}{\\lim}\\frac{(2-x)}{(x-2)(2x)} & & & \\text{Simplify.} \\\\ & =\\underset{x\\to 2}{\\lim}\\frac{-1}{2x} & & & \\text{Simplify using} \\, \\frac{2-x}{x-2}=-1, \\, \\text{for} \\, x\\ne 2. \\\\ & =-\\frac{1}{4} & & & \\text{Evaluate the limit.} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169739270491\">We now know that the slope of the tangent line is [latex]-\\frac{1}{4}[\/latex]. To find the equation of the tangent line, we also need a point on the line. We know that [latex]f(2)=\\frac{1}{2}[\/latex]. Since the tangent line passes through the point [latex](2,\\frac{1}{2})[\/latex] we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation [latex]y=-\\frac{1}{4}x+1[\/latex]. The graphs of [latex]f(x)=\\frac{1}{x}[\/latex] and [latex]y=-\\frac{1}{4}x+1[\/latex] are shown in Figure 6.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205200\/CNX_Calc_Figure_03_01_006.jpg\" alt=\"This figure consists of the graphs of f(x) = 1\/x and y = -x\/4 + 1. The part of the graph f(x) = 1\/x in the first quadrant appears to touch the other function\u2019s graph at x = 2.\" width=\"487\" height=\"321\" \/><figcaption class=\"wp-caption-text\">Figure 6. The line is tangent to [latex]f(x)[\/latex] at [latex]x=2[\/latex]<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739231715\">Find the slope of the line tangent to the graph of [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=4[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q3776221\">Hint<\/button><\/p>\n<div id=\"q3776221\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738973750\">Use either definition. Multiply the numerator and the denominator by a conjugate.<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739236594\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739236594\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739236594\">[latex]\\dfrac{1}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to this example. <br \/>\n<iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=738&amp;end=892&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.1DefiningTheDerivative738to892_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Defining the Derivative&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find the equation of the line tangent to [latex]f(x) = x^2[\/latex] at [latex]x = 3[\/latex].<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q948182\">Show Answer<\/button><\/p>\n<div id=\"q948182\" class=\"hidden-answer\" style=\"display: none\">\n<p>Calculate the slope:<\/p>\n<p style=\"text-align: center;\">[latex]<br \/>  \\begin{array}{rcl}<br \/>  m_{\\text{tan}} &=& \\lim_{x \\to 3} \\frac{f(x) - f(3)}{x - 3} \\\\<br \/>  &=& \\lim_{x \\to 3} \\frac{x^2 - 9}{x - 3} \\\\<br \/>  &=& \\lim_{x \\to 3} (x + 3) \\\\<br \/>  &=& 6<br \/>  \\end{array}<br \/>[\/latex]<\/p>\n<p>Identify the point:<\/p>\n<p style=\"text-align: center;\">[latex](3, f(3)) = (3, 9)[\/latex]<\/p>\n<p>Use point-slope form:<\/p>\n<p style=\"text-align: center;\">[latex]y - 9 = 6(x - 3)[\/latex]<\/p>\n<p>Simplify to slope-intercept form:<\/p>\n<p style=\"text-align: center;\">[latex]y = 6x - 9[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<h2>The Derivative of a Function at a Point<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Definition of the Derivative:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">The derivative is the instantaneous rate of change of a function at a point<\/li>\n<li class=\"whitespace-normal break-words\">It&#8217;s equivalent to the slope of the tangent line at that point<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Notations for the Derivative:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]f'(a)[\/latex] denotes the derivative of [latex]f(x)[\/latex] at [latex]x = a[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Two Equivalent Definitions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]f'(a) = \\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f'(a) = \\lim_{h \\to 0} \\frac{f(a+h) - f(a)}{h}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>Formal definition of a derivative:\n<p class=\"whitespace-pre-wrap break-words\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0The derivative of a function [latex]f(x)[\/latex] at a point [latex]a[\/latex] is defined as:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]f'(a) = \\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0or equivalently:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]f'(a) = \\lim_{h \\to 0} \\frac{f(a+h) - f(a)}{h}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0provided these limits exist.<\/p>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739001692\">For [latex]f(x)=3x^2-4x+1[\/latex], find [latex]f^{\\prime}(2)[\/latex] by using the first definition.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739104712\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739104712\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739104712\">Substitute the given function and value directly into the equation.<\/p>\n<div id=\"fs-id1169739305050\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}f^{\\prime}(x)& =\\underset{x\\to 2}{\\lim}\\frac{f(x)-f(2)}{x-2} & & & \\text{Apply the definition.} \\\\ & =\\underset{x\\to 2}{\\lim}\\frac{(3x^2-4x+1)-5}{x-2} & & & \\text{Substitute} \\, f(x)=3x^2-4x+1 \\, \\text{and} \\, f(2)=5. \\\\ & =\\underset{x\\to 2}{\\lim}\\frac{(x-2)(3x+2)}{x-2} & & & \\text{Simplify and factor the numerator.} \\\\ & =\\underset{x\\to 2}{\\lim}(3x+2) & & & \\text{Cancel the common factor.} \\\\ & =8 & & & \\text{Evaluate the limit.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739269938\">For [latex]f(x)=x^2+3x+2[\/latex], find [latex]f^{\\prime}(1)[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q708365\">Hint<\/button><\/p>\n<div id=\"q708365\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736618844\">Use either the first definition, the second, or try both.\u00a0<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739293550\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739293550\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739293550\">[latex]f^{\\prime}(1)=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=1434&amp;end=1537&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.1DefiningTheDerivative1434to1537_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Defining the Derivative&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Find [latex]f'(2)[\/latex] for [latex]f(x) = 3x^2 - 4x + 1[\/latex] using the second definition.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q723055\">Show Answer<\/button><\/p>\n<div id=\"q723055\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\begin{array}{rcl} f'(2) &=& \\lim_{h \\to 0} \\frac{f(2+h) - f(2)}{h} \\\\ &=& \\lim_{h \\to 0} \\frac{[3(2+h)^2 - 4(2+h) + 1] - [3(2)^2 - 4(2) + 1]}{h} \\\\ &=& \\lim_{h \\to 0} \\frac{[12 + 12h + 3h^2 - 8 - 4h + 1] - [12 - 8 + 1]}{h} \\\\ &=& \\lim_{h \\to 0} \\frac{3h^2 + 8h}{h} \\\\ &=& \\lim_{h \\to 0} (3h + 8) \\\\ &=& 8 \\end{array}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore, [latex]f'(2) = 8[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<h2>Velocities and Rates of Change<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Average Velocity:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For a position function [latex]s(t)[\/latex], average velocity over [latex][a,t][\/latex] is: [latex]v_{\\text{avg}} = \\frac{s(t) - s(a)}{t - a}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Instantaneous Velocity:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Limit of average velocity as time interval approaches zero<\/li>\n<li class=\"whitespace-normal break-words\">Equivalent to the derivative of the position function: [latex]v(a) = s'(a) = \\lim_{t \\to a} \\frac{s(t) - s(a)}{t - a}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Instantaneous Rate of Change:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Generalization of instantaneous velocity to any function<\/li>\n<li class=\"whitespace-normal break-words\">For a function [latex]f(x)[\/latex], the instantaneous rate of change at [latex]a[\/latex] is [latex]f'(a)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Graphical Interpretation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Average velocity: Slope of secant line<\/li>\n<li class=\"whitespace-normal break-words\">Instantaneous velocity: Slope of tangent line<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739220848\">A rock is dropped from a height of [latex]64[\/latex] feet. Its height above ground at time [latex]t[\/latex] seconds later is given by [latex]s(t)=-16t^2+64, \\, 0\\le t\\le 2[\/latex]. Find its instantaneous velocity [latex]1[\/latex] second after it is dropped, using the definition of a derivative.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q403817\">Hint<\/button><\/p>\n<div id=\"q403817\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739189236\">[latex]v(t)=s^{\\prime}(t)[\/latex].\u00a0<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739298454\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739298454\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739298454\">[latex]-32[\/latex] ft\/sec<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=1660&amp;end=1763&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.1DefiningTheDerivative1660to1763_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Defining the Derivative&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739253553\">A toy company can sell [latex]x[\/latex] electronic gaming systems at a price of [latex]p=-0.01x+400[\/latex] dollars per gaming system. The cost of manufacturing [latex]x[\/latex] systems is given by [latex]C(x)=100x+10,000[\/latex] dollars. Find the rate of change of profit when [latex]10,000[\/latex] games are produced. Should the toy company increase or decrease production?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739274608\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739274608\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739274608\">The profit [latex]P(x)[\/latex] earned by producing [latex]x[\/latex] gaming systems is [latex]R(x)-C(x)[\/latex], where [latex]R(x)[\/latex] is the revenue obtained from the sale of [latex]x[\/latex] games. Since the company can sell [latex]x[\/latex] games at [latex]p=-0.01x+400[\/latex] per game,<\/p>\n<div id=\"fs-id1169739286461\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp=x(-0.01x+400)=-0.01x^2+400x[\/latex].<\/div>\n<p id=\"fs-id1169739302924\">Consequently,<\/p>\n<div id=\"fs-id1169739302928\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(x)=-0.01x^2+300x-10,000[\/latex].<\/div>\n<p id=\"fs-id1169739190488\">Therefore, evaluating the rate of change of profit gives<\/p>\n<div id=\"fs-id1169739190491\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}P^{\\prime}(10000)& =\\underset{x\\to 10000}{\\lim}\\frac{P(x)-P(10000)}{x-10000} \\\\ & =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-10000-1990000}{x-10000} \\\\ & =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-2000000}{x-10000} \\\\ & =100 \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169736587942\">Since the rate of change of profit [latex]P^{\\prime}(10,000)>0[\/latex] and [latex]P(10,000)>0[\/latex], the company should increase production.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":503,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2817"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":14,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2817\/revisions"}],"predecessor-version":[{"id":4680,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2817\/revisions\/4680"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/503"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2817\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2817"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2817"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2817"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2817"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}