3.2 The derivative as a function  (Page 3/10)

Proof

If $f(x)$ is differentiable at $a,$ then $f^{\prime }(a)$ exists and

We want to show that $f(x)$ is continuous at $a$ by showing that $\underset{x\to a}{\text{lim}}f(x)=f(a).$ Thus,

Therefore, since $f\left(a\right)$ is defined and $\underset{x\to a}{\text{lim}}f\left(x\right)=f(a),$ we conclude that $f$ is continuous at $a.$

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We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the function $f\left(x\right)=\left|x\right|.$ This function is continuous everywhere; however, $f^{\prime }(0)$ is undefined. This observation leads us to believe that continuity does not imply differentiability. Let’s explore further. For $f\left(x\right)=\left|x\right|,$

This limit does not exist because

See [link] .

Let’s consider some additional situations in which a continuous function fails to be differentiable. Consider the function $f\left(x\right)=\sqrt[3]{x}\text{:}$

Thus $f^{\prime }\left(0\right)$ does not exist. A quick look at the graph of $f\left(x\right)=\sqrt[3]{x}$ clarifies the situation. The function has a vertical tangent line at $0$ ( [link] ).

The function $f\left(x\right)=\{\begin{array}{l}x\text{sin}\left(\frac{1}{x}\right)\text{if}x\ne 0\\ 0\text{if}x=0\end{array}$ also has a derivative that exhibits interesting behavior at $0.$ We see that

This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero ( [link] ).

In summary:

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.
2. We saw that $f\left(x\right)=\left|x\right|$ failed to be differentiable at $0$ 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 $0.$ From this we conclude that in order to be differentiable at a point, a function must be “smooth” at that point.
3. As we saw in the example of $f\left(x\right)=\sqrt[3]{x},$ a function fails to be differentiable at a point where there is a vertical tangent line.
4. As we saw with $f\left(x\right)=\{\begin{array}{l}x\text{sin}\left(\frac{1}{x}\right)\text{if}x\ne 0\\ 0\text{if}x=0\end{array}$ a function may fail to be differentiable at a point in more complicated ways as well.

A piecewise function that is continuous and differentiable

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 ( [link] ). The function that describes the track is to have the form $f\left(x\right)=\{\begin{array}{c}\frac{1}{10}{x}^2+bx+c\text{if}x<\mathrm{-10}\\ -\frac{1}{4}x+\frac{5}{2}\text{if}x\ge \mathrm{-10}\end{array}$ where $x$ and $f(x)$ are in inches. For the car to move smoothly along the track, the function $f(x)$ must be both continuous and differentiable at $\mathrm{-10}.$ Find values of $b$ and $c$ that make $f(x)$ both continuous and differentiable.

For the function to be continuous at $x=\mathrm{-10},\underset{x\to {10}^{-}}{\text{lim}}f\left(x\right)=f\left(\mathrm{-10}\right).$ Thus, since

$\underset{x\to \text{−}{10}^{-}}{\text{lim}}f\left(x\right)=\frac{1}{10}{(\mathrm{-10})}^2-10b+c=10-10b+c$

and $f\left(\mathrm{-10}\right)=5,$ we must have $10-10b+c=5.$ Equivalently, we have $c=10b-5.$

For the function to be differentiable at $\mathrm{-10},$

$f^{\prime }\left(10\right)=\underset{x\to \text{−}10}{\text{lim}}\frac{f\left(x\right)-f(\mathrm{-10})}{x+10}$

must exist. Since $f\left(x\right)$ 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:

$\begin{array}{ccccc}\hfill \underset{x\to \text{−}{10}^{-}}{\text{lim}}\frac{f\left(x\right)-f(\mathrm{-10})}{x+10}& =\underset{x\to \text{−}{10}^{-}}{\text{lim}}\frac{\frac{1}{10}{x}^2+bx+c-5}{x+10}\hfill & & & \\ & =\underset{x\to \text{−}{10}^{-}}{\text{lim}}\frac{\frac{1}{10}{x}^2+bx+\left(10b-5\right)-5}{x+10}\hfill & & & \text{Substitute}c=10b-5.\hfill \\ & =\underset{x\to \text{−}{10}^{-}}{\text{lim}}\frac{x^2-100+10bx+100b}{10(x+10)}\hfill & & & \\ & =\underset{x\to \text{−}{10}^{-}}{\text{lim}}\frac{(x+10)(x-10+10b)}{10(x+10)}\hfill & & & \text{Factor by grouping.}\hfill \\ & =b-2.\hfill & & & \end{array}$

We also have

$\begin{array}{cc}\hfill \underset{x\to \text{−}{10}^{+}}{\text{lim}}\frac{f\left(x\right)-f(\mathrm{-10})}{x+10}& =\underset{x\to \text{−}{10}^{+}}{\text{lim}}\frac{-\frac{1}{4}x+\frac{5}{2}-5}{x+10}\hfill \\ & =\underset{x\to \text{−}{10}^{+}}{\text{lim}}\frac{\text{−}(x+10)}{4(x+10)}\hfill \\ & =-\frac{1}{4}.\hfill \end{array}$

This gives us $b-2=-\frac{1}{4}.$ Thus $b=\frac{7}{4}$ and $c=10\left(\frac{7}{4}\right)-5=\frac{25}{2}.$

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