Derivatives and Continuity

Now that we can graph a derivative, let’s examine the behavior of the graphs.

First, 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.

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 [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’s explore further.

For [latex]f(x)=|x|[/latex],

[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]

This limit does not exist because

[latex]\underset{x\to 0^-}{\lim}\dfrac{|x|}{x}=-1 \, \text{and} \, \underset{x\to 0^+}{\lim}\dfrac{|x|}{x}=1[/latex]

Let’s consider some additional situations in which a continuous function fails to be differentiable.

Consider the function [latex]f(x)=\sqrt[3]{x}[/latex]:

[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]

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).

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].

We see that,

[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]

This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure 8).

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 [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 “smooth” at that point.
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.
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.

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

[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],

where [latex]x[/latex] and [latex]f(x)[/latex] are in inches.

For 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.

Show Solution

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

[latex]\underset{x\to −10^-}{\lim}f(x)=\frac{1}{10}(-10)^2-10b+c=10-10b+c[/latex]

and [latex]f(-10)=5[/latex], we must have [latex]10-10b+c=5[/latex]. Equivalently, we have [latex]c=10b-5[/latex].

For the function to be differentiable at [latex]-10[/latex],

[latex]f^{\prime}(10)=\underset{x\to −10}{\lim}\frac{f(x)-f(-10)}{x+10}[/latex]

must exist.

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:

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

We also have

[latex]\begin{array}{ll} \underset{x\to −10^+}{\lim}\frac{f(x)-f(-10)}{x+10} & =\underset{x\to −10^+}{\lim}\frac{-\frac{1}{4}x+\frac{5}{2}-5}{x+10} \\ & =\underset{x\to −10^+}{\lim}\frac{−(x+10)}{4(x+10)} \\ & =-\frac{1}{4} \end{array}[/latex]

This gives us [latex]b-2=-\frac{1}{4}[/latex].

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