[latex]f^{\prime}(x)=2x[/latex]

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.2 The Derivative as a Function” here (opens in new window).

[latex]\begin{array}{rcl} f'(x) &=& \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &=& \lim_{h \to 0} \frac{((x+h)^3 + 2(x+h)) - (x^3 + 2x)}{h} \\ &=& \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3 + 2x + 2h) - (x^3 + 2x)}{h} \\ &=& \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 + 2h}{h} \\ &=& \lim_{h \to 0} (3x^2 + 3xh + h^2 + 2) \\ &=& 3x^2 + 2 \end{array}[/latex]

Therefore, [latex]f'(x) = 3x^2 + 2[/latex].

The graph of [latex]f^{\prime}(x)[/latex] is positive where [latex]f(x)[/latex] is increasing.

[latex](0,+\infty)[/latex]

The key points are:

• [latex]f'(0) = 0[/latex]
• [latex]f'(2) = 0[/latex] (local maximum of [latex]f(x)[/latex])
• [latex]f'(1) = -3[/latex] (inflection point of [latex]f(x)[/latex])

Analyzing the intervals:

• [latex]f'(x) < 0[/latex] for [latex]0 < x < 2[/latex] ([latex]f(x)[/latex] decreasing)
• [latex]f'(x) > 0[/latex] for [latex]x < 0[/latex] and [latex]x > 2[/latex] ([latex]f(x)[/latex] increasing)

Our sketch of [latex]f'(x)[/latex] should include:

• Parabola opening upward
• [latex]x[/latex]-intercepts at [latex]x = 0[/latex] and [latex]x = 2[/latex]
• Vertex at [latex](1, -3)[/latex]

For continuity at [latex]x = 2[/latex]:

[latex]4a + 2b = 6[/latex]

For differentiability at [latex]x = 2[/latex]:

[latex]2ax + b = 1[/latex] when [latex]x = 2[/latex]
[latex]4a + b = 1[/latex]

Solve the system of equations:

[latex]4a + 2b = 6[/latex]
[latex]4a + b = 1[/latex]

Subtracting the second equation from the first:

[latex]b = 5[/latex]

Substituting back:

[latex]4a + 5 = 1[/latex]
[latex]4a = -4[/latex]
[latex]a = -1[/latex]

Therefore, [latex]a = -1[/latex] and [latex]b = 5[/latex] make the function both continuous and differentiable at [latex]x = 2[/latex].

Check continuity at [latex]x = 0[/latex]:

[latex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0[/latex]
[latex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sqrt{x} = 0[/latex]
[latex]f(0) = 0^2 = 0[/latex]

The function is continuous at [latex]x = 0[/latex].

Check differentiability at [latex]x = 0[/latex]:

Left-hand derivative:

[latex]\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h^2 - 0}{h} = \lim_{h \to 0^-} h = 0[/latex]

Right-hand derivative:

[latex]\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\sqrt{h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{\sqrt{h}} = \infty[/latex]

The left-hand and right-hand derivatives are not equal, so [latex]f(x)[/latex] is not differentiable at [latex]x = 0[/latex].

Conclusion: The function is continuous everywhere but not differentiable at [latex]x = 0[/latex].

The absolute value function can create points of non-differentiability where the expression inside changes sign.

Set [latex]x^2 - 4x + 3 = 0[/latex]:

[latex](x - 1)(x - 3) = 0[/latex]
[latex]x = 1[/latex] or [latex]x = 3[/latex]

Check differentiability at [latex]x = 1[/latex]:

Left-hand derivative:

[latex]\lim_{h \to 0^-} \frac{|((1+h)^2 - 4(1+h) + 3)| - 0}{h} = \lim_{h \to 0^-} \frac{|-h^2 - 2h|}{h} = -2[/latex]

Right-hand derivative:

[latex]\lim_{h \to 0^+} \frac{|((1+h)^2 - 4(1+h) + 3)| - 0}{h} = \lim_{h \to 0^+} \frac{|-h^2 - 2h|}{h} = 2[/latex]

The function is not differentiable at [latex]x = 1[/latex].

Check differentiability at [latex]x = 3[/latex]:

Left-hand derivative:

[latex]\lim_{h \to 0^-} \frac{|((3+h)^2 - 4(3+h) + 3)| - 0}{h} = \lim_{h \to 0^-} \frac{|h^2 + 2h|}{h} = 2[/latex]

Right-hand derivative:

[latex]\lim_{h \to 0^+} \frac{|((3+h)^2 - 4(3+h) + 3)| - 0}{h} = \lim_{h \to 0^+} \frac{|h^2 + 2h|}{h} = 2[/latex]

The function is differentiable at [latex]x = 3[/latex].

Conclusion: The function is not differentiable at [latex]x = 1[/latex], but is differentiable at all other points.

[latex]f''(x)=2[/latex]

Watch the following video to see the worked solution to this example.

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.

You can view the transcript for this segmented clip of “3.2 The Derivative as a Function” here (opens in new window).

First derivative:

[latex]f'(x) = 4x^3 - 6x^2 + 6x - 4[/latex]

Second derivative:

[latex]f''(x) = 12x^2 - 12x + 6[/latex]

Third derivative:

[latex]f'''(x) = 24x - 12[/latex]

Fourth derivative:

[latex]f^{(4)}(x) = 24[/latex]

All subsequent derivatives will be zero.

[latex]a(t)=6t[/latex]