{"id":2841,"date":"2024-06-10T17:53:27","date_gmt":"2024-06-10T17:53:27","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=2841"},"modified":"2024-08-05T12:38:50","modified_gmt":"2024-08-05T12:38:50","slug":"the-derivative-as-a-function-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-derivative-as-a-function-learn-it-3\/","title":{"raw":"The Derivative as a Function: Learn It 3","rendered":"The Derivative as a Function: Learn It 3"},"content":{"raw":"

Derivatives and Continuity<\/h2>\r\n

Now that we can graph a derivative, let\u2019s examine the behavior of the graphs.
\r\n
\r\nFirst, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons.<\/p>\r\n

\r\n

differentiability implies continuity<\/h3>\r\n

Let [latex]f(x)[\/latex] be a function and [latex]a[\/latex] be in its domain. If [latex]f(x)[\/latex] is differentiable at [latex]a[\/latex], then [latex]f[\/latex] is continuous at [latex]a[\/latex].<\/p>\r\n<\/section>\r\n

\r\n

Proof<\/strong><\/p>\r\n

\r\n

If [latex]f(x)[\/latex] is differentiable at [latex]a[\/latex], then [latex]f^{\\prime}(a)[\/latex] exists and<\/p>\r\n

[latex]f^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(x)-f(a)}{x-a}[\/latex]<\/div>\r\n

 <\/p>\r\n

We want to show that [latex]f(x)[\/latex] is continuous at [latex]a[\/latex] by showing that [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex]. Thus,<\/p>\r\n

[latex]\\begin{array}{lllll} \\underset{x\\to a}{\\lim}f(x) & =\\underset{x\\to a}{\\lim}(f(x)-f(a)+f(a)) & & & \\\\ & =\\underset{x\\to a}{\\lim}(\\frac{f(x)-f(a)}{x-a}\\cdot (x-a)+f(a)) & & & \\text{Multiply and divide} \\, f(x)-f(a) \\, \\text{by} \\, x-a. \\\\ & =(\\underset{x\\to a}{\\lim}\\frac{f(x)-f(a)}{x-a}) \\cdot (\\underset{x\\to a}{\\lim}(x-a))+\\underset{x\\to a}{\\lim}f(a) & & & \\\\ & =f^{\\prime}(a) \\cdot 0+f(a) & & & \\\\ & =f(a). & & & \\end{array}[\/latex]<\/div>\r\n

 <\/p>\r\n

Therefore, since [latex]f(a)[\/latex] is defined and [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex], we conclude that [latex]f[\/latex] is continuous at [latex]a[\/latex].<\/p>\r\n

[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n

We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability.
\r\n
\r\nTo determine an answer to this question, we examine the function [latex]f(x)=|x|[\/latex]. This function is continuous everywhere; however, [latex]f^{\\prime}(0)[\/latex] is undefined. This observation leads us to believe that continuity does not imply differentiability. Let\u2019s explore further.<\/p>\r\n

\r\n

For [latex]f(x)=|x|[\/latex],<\/p>\r\n

[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{f(x)-f(0)}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{|x|-|0|}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{|x|}{x}[\/latex]<\/div>\r\n

This limit does not exist because<\/p>\r\n

[latex]\\underset{x\\to 0^-}{\\lim}\\dfrac{|x|}{x}=-1 \\, \\text{and} \\, \\underset{x\\to 0^+}{\\lim}\\dfrac{|x|}{x}=1[\/latex]<\/div>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"The Figure 6. The function [latex]f(x)=|x|[\/latex] is continuous at 0 but is not differentiable at 0.[\/caption]\r\n<\/section>\r\n

Let\u2019s consider some additional situations in which a continuous function fails to be differentiable.<\/p>\r\n

\r\n

Consider the function [latex]f(x)=\\sqrt[3]{x}[\/latex]:<\/p>\r\n

[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{\\sqrt[3]{x}-0}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{1}{\\sqrt[3]{x^2}}=+\\infty[\/latex]<\/div>\r\n

Thus [latex]f^{\\prime}(0)[\/latex] does not exist. A quick look at the graph of [latex]f(x)=\\sqrt[3]{x}[\/latex] clarifies the situation. The function has a vertical tangent line at [latex]0[\/latex] (Figure 7).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"The Figure 7. The function [latex]f(x)=\\sqrt[3]{x}[\/latex] has a vertical tangent at [latex]x=0[\/latex]. It is continuous at 0 but is not differentiable at 0.[\/caption]\r\n<\/section>\r\n

The function [latex]f(x)=\\begin{cases} x \\sin\\left(\\frac{1}{x}\\right) & \\text{ if } \\, x \\ne 0 \\\\ 0 & \\text{ if } \\, x = 0 \\end{cases}[\/latex] also has a derivative that exhibits interesting behavior at [latex]0[\/latex].<\/p>\r\n

\r\n

We see that,<\/p>\r\n

[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{x \\sin\\left(\\frac{1}{x}\\right)-0}{x-0}=\\underset{x\\to 0}{\\lim} \\sin\\left(\\dfrac{1}{x}\\right)[\/latex]<\/div>\r\n

This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure 8).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"426\"]\"The Figure 8. The function [latex]f(x)=\\begin{cases} x \\sin(\\frac{1}{x}) & \\text{ if } \\, x \\ne 0 \\\\ 0 & \\text{ if } \\, x = 0 \\end{cases}[\/latex] is not differentiable at 0.[\/caption]\r\n<\/section>\r\n

In summary:<\/p>\r\n

    \r\n\t
  1. We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable.<\/li>\r\n\t
  2. We saw that [latex]f(x)=|x|[\/latex] failed to be differentiable at [latex]0[\/latex] because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at [latex]0[\/latex]. From this we conclude that in order to be differentiable at a point, a function must be \u201csmooth\u201d at that point.<\/li>\r\n\t
  3. As we saw in the example of [latex]f(x)=\\sqrt[3]{x}[\/latex], a function fails to be differentiable at a point where there is a vertical tangent line.<\/li>\r\n\t
  4. As we saw with [latex]f(x)=\\begin{cases} x \\sin(\\frac{1}{x}) & \\text{ if } \\, x \\ne 0 \\\\ 0 & \\text{ if } \\, x = 0 \\end{cases}[\/latex] a function may fail to be differentiable at a point in more complicated ways as well.<\/li>\r\n<\/ol>\r\n
    \r\n

    A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts to a straight line (Figure 9). The function that describes the track is to have the form<\/p>\r\n

    [latex]f(x)=\\begin{cases} \\frac{1}{10}x^2 + bx + c & \\text{ if } \\, x < -10 \\\\ -\\frac{1}{4}x + \\frac{5}{2} & \\text{ if } \\, x \\ge -10 \\end{cases}[\/latex],<\/center>\r\n

    where [latex]x[\/latex] and [latex]f(x)[\/latex] are in inches.
    \r\n
    \r\nFor the car to move smoothly along the track, the function [latex]f(x)[\/latex] must be both continuous and differentiable at [latex]-10[\/latex]. Find values of [latex]b[\/latex] and [latex]c[\/latex] that make [latex]f(x)[\/latex] both continuous and differentiable.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 9. For the car to move smoothly along the track, the function must be both continuous and differentiable.[\/caption]\r\n\r\n\r\n

    [reveal-answer q=\"fs-id1169738222036\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"fs-id1169738222036\"]
    \r\n
    \r\n<\/p>\r\n

    For the function to be continuous at [latex]x=-10, \\, \\underset{x\\to 10^-}{\\lim}f(x)=f(-10)[\/latex]. Thus, since<\/p>\r\n

    [latex]\\underset{x\\to \u221210^-}{\\lim}f(x)=\\frac{1}{10}(-10)^2-10b+c=10-10b+c[\/latex]<\/div>\r\n

    and [latex]f(-10)=5[\/latex], we must have [latex]10-10b+c=5[\/latex]. Equivalently, we have [latex]c=10b-5[\/latex].<\/p>\r\n

    For the function to be differentiable at [latex]-10[\/latex],<\/p>\r\n

    [latex]f^{\\prime}(10)=\\underset{x\\to \u221210}{\\lim}\\frac{f(x)-f(-10)}{x+10}[\/latex]<\/div>\r\n

    must exist.<\/p>\r\n

    Since [latex]f(x)[\/latex] is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:<\/p>\r\n

    [latex]\\begin{array}{lllll} \\underset{x\\to \u221210^-}{\\lim}\\frac{f(x)-f(-10)}{x+10} & =\\underset{x\\to \u221210^-}{\\lim}\\frac{\\frac{1}{10}x^2+bx+c-5}{x+10} & & & \\\\ & =\\underset{x\\to \u221210^-}{\\lim}\\frac{\\frac{1}{10}x^2+bx+(10b-5)-5}{x+10} & & & \\text{Substitute} \\, c=10b-5. \\\\ & =\\underset{x\\to \u221210^-}{\\lim}\\frac{x^2-100+10bx+100b}{10(x+10)} & & & \\\\ & =\\underset{x\\to \u221210^-}{\\lim}\\frac{(x+10)(x-10+10b)}{10(x+10)} & & & \\text{Factor by grouping.} \\\\ & =b-2 & & & \\end{array}[\/latex]<\/div>\r\n

    We also have<\/p>\r\n

    [latex]\\begin{array}{ll} \\underset{x\\to \u221210^+}{\\lim}\\frac{f(x)-f(-10)}{x+10} & =\\underset{x\\to \u221210^+}{\\lim}\\frac{-\\frac{1}{4}x+\\frac{5}{2}-5}{x+10} \\\\ & =\\underset{x\\to \u221210^+}{\\lim}\\frac{\u2212(x+10)}{4(x+10)} \\\\ & =-\\frac{1}{4} \\end{array}[\/latex]<\/div>\r\n

    This gives us [latex]b-2=-\\frac{1}{4}[\/latex].<\/p>\r\n

    Thus [latex]b=\\frac{7}{4}[\/latex] and [latex]c=10(\\frac{7}{4})-5=\\frac{25}{2}[\/latex].<\/p>\r\n

    Watch the following video to see the worked solution to this example.<\/p>\r\n