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.

Theorem 3.1

Differentiability Implies Continuity

Let \(f \left(\right. x \left.\right)\) be a function and \(a\) be in its domain. If \(f \left(\right. x \left.\right)\) is differentiable at \(a ,\) then \(f\) is continuous at \(a .\)

Proof

If \(f \left(\right. x \left.\right)\) is differentiable at \(a ,\) then \(f^{'} \left(\right. a \left.\right)\) exists and

\[f^{'} \left(\right. a \left.\right) = \underset{x \rightarrow a}{\text{lim}} \frac{f \left(\right. x \left.\right) - f \left(\right. a \left.\right)}{x - a} .\]

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

\[\begin{aligned} \underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) & = \underset{x \rightarrow a}{\text{lim}} \left(\right. f \left(\right. x \left.\right) - f \left(\right. a \left.\right) + f \left(\right. a \left.\right) \left.\right) & & & \\ & = \underset{x \rightarrow a}{\text{lim}} \left(\right. \frac{f \left(\right. x \left.\right) - f \left(\right. a \left.\right)}{x - a} \cdot \left(\right. x - a \left.\right) + f \left(\right. a \left.\right) \left.\right) & & & \text{Multiply and divide} f \left(\right. x \left.\right) - f \left(\right. a \left.\right) \text{by} x - a . \\ & = \left(\right. \underset{x \rightarrow a}{\text{lim}} \frac{f \left(\right. x \left.\right) - f \left(\right. a \left.\right)}{x - a} \left.\right) \cdot \left(\right. \underset{x \rightarrow a}{\text{lim}} \left(\right. x - a \left.\right) \left.\right) + \underset{x \rightarrow a}{\text{lim}} f \left(\right. a \left.\right) & & & \\ & = f ' \left(\right. a \left.\right) \cdot 0 + f \left(\right. a \left.\right) & & & \\ & = f \left(\right. a ). & & & \end{aligned}\]

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

□

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 \(f \left(\right. x \left.\right) = \left|\right. x \left|\right. .\) This function is continuous everywhere; however, \(f^{'} \left(\right. 0 \left.\right)\) is undefined. This observation leads us to believe that continuity does not imply differentiability. Let’s explore further. For \(f \left(\right. x \left.\right) = \left|\right. x \left|\right. ,\)

\[f^{'} \left(\right. 0 \left.\right) = \underset{x \rightarrow 0}{\text{lim}} \frac{f \left(\right. x \left.\right) - f \left(\right. 0 \left.\right)}{x - 0} = \underset{x \rightarrow 0}{\text{lim}} \frac{\left|\right. x \left|\right. - \left|\right. 0 \left|\right.}{x - 0} = \underset{x \rightarrow 0}{\text{lim}} \frac{\left|\right. x \left|\right.}{x} .\]

This limit does not exist because

\[\underset{x \rightarrow 0^{-}}{\text{lim}} \frac{\left|\right. x \left|\right.}{x} = −1 \text{and} \underset{x \rightarrow 0^{+}}{\text{lim}} \frac{\left|\right. x \left|\right.}{x} = 1 .\]

See Figure 3.14.

The function f(x) = the absolute value of x is graphed. It consists of two straight line segments: the first follows the equation y = −x and ends at the origin; the second follows the equation y = x and starts at the origin.

Figure 3.14 The function \(f \left(\right. x \left.\right) = \left|\right. x \left|\right.\) is continuous at \(0\) but is not differentiable at \(0 .\)

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

\[f^{'} \left(\right. 0 \left.\right) = \underset{x \rightarrow 0}{\text{lim}} \frac{\sqrt[3]{x} - 0}{x - 0} = \underset{x \rightarrow 0}{\text{lim}} \frac{1}{\sqrt[3]{x^{2}}} = + \infty .\]

Thus \(f^{'} \left(\right. 0 \left.\right)\) does not exist. A quick look at the graph of \(f \left(\right. x \left.\right) = \sqrt[3]{x}\) clarifies the situation. The function has a vertical tangent line at \(0\) (Figure 3.15).

The function f(x) = the cube root of x is graphed. It has a vertical tangent at x = 0.

Figure 3.15 The function \(f \left(\right. x \left.\right) = \sqrt[3]{x}\) has a vertical tangent at \(x = 0 .\) It is continuous at \(0\) but is not differentiable at \(0 .\)

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

\[f^{'} \left(\right. 0 \left.\right) = \underset{x \rightarrow 0}{\text{lim}} \frac{x \text{sin} \left(\right. 1 / x \left.\right) - 0}{x - 0} = \underset{x \rightarrow 0}{\text{lim}} \text{sin} \left(\right. \frac{1}{x} \left.\right) .\]

This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure 3.16).

The function f(x) = x sin (1/2) if x does not equal 0 and f(x) = 0 if x = 0 is graphed. It looks like a rapidly oscillating sinusoidal function with amplitude decreasing to 0 at the origin.

Figure 3.16 The function \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. x \text{sin} \left(\right. \frac{1}{x} \left.\right) \text{if} x \neq 0 \\ 0 \text{if} x = 0 \end{aligned}\) is not differentiable at \(0 .\)

In summary:

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.
We saw that \(f \left(\right. x \left.\right) = \left|\right. x \left|\right.\) 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.
As we saw in the example of \(f \left(\right. x \left.\right) = \sqrt[3]{x} ,\) a function fails to be differentiable at a point where there is a vertical tangent line.
As we saw with \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. x \text{sin} \left(\right. \frac{1}{x} \left.\right) \text{if} x \neq 0 \\ 0 \text{if} x = 0 \end{aligned}\) a function may fail to be differentiable at a point in more complicated ways as well.

Example 3.14

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 (Figure 3.17). The function that describes the track is to have the form \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. \frac{1}{10} x^{2} + b x + c \text{if} x < −10 \\ - \frac{1}{4} x + \frac{5}{2} \text{if} x \geq −10 \end{aligned}\) where \(x\) and \(f \left(\right. x \left.\right)\) are in inches. For the car to move smoothly along the track, the function \(f \left(\right. x \left.\right)\) must be both continuous and differentiable at \(−10 .\) Find values of \(b\) and \(c\) that make \(f \left(\right. x \left.\right)\) both continuous and differentiable.

A cart is drawn on a line that curves through (−10, 5) to (10, 0) with y-intercept roughly (0, 2).

Figure 3.17 For the car to move smoothly along the track, the function must be both continuous and differentiable.

Solution

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

\[\underset{x \rightarrow − 10^{-}}{\text{lim}} f \left(\right. x \left.\right) = \frac{1}{10} \left(\left(\right. −10 \left.\right)\right)^{2} - 10 b + c = 10 - 10 b + c\]

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

For the function to be differentiable at \(−10 ,\)

\[f^{'} \left(\right. -10 \left.\right) = \underset{x \rightarrow − 10}{\text{lim}} \frac{f \left(\right. x \left.\right) - f \left(\right. −10 \left.\right)}{x + 10}\]

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

\[\begin{aligned} \underset{x \rightarrow − 10^{-}}{\text{lim}} \frac{f \left(\right. x \left.\right) - f \left(\right. −10 \left.\right)}{x + 10} & = \underset{x \rightarrow − 10^{-}}{\text{lim}} \frac{\frac{1}{10} x^{2} + b x + c - 5}{x + 10} & & & \\ & = \underset{x \rightarrow − 10^{-}}{\text{lim}} \frac{\frac{1}{10} x^{2} + b x + \left(\right. 10 b - 5 \left.\right) - 5}{x + 10} & & & \text{Substitute} c = 10 b - 5. \\ & = \underset{x \rightarrow − 10^{-}}{\text{lim}} \frac{x^{2} - 100 + 10 b x + 100 b}{10 \left(\right. x + 10 \left.\right)} & & & \text{Multiply numerator and denominator by 10}. \\ & = \underset{x \rightarrow − 10^{-}}{\text{lim}} \frac{\left(\right. x + 10 \left.\right) \left(\right. x - 10 + 10 b \left.\right)}{10 \left(\right. x + 10 \left.\right)} & & & \text{Factor by grouping}. \\ & = b - 2. & & & \end{aligned}\]

We also have

\[\begin{aligned} \underset{x \rightarrow − 10^{+}}{\text{lim}} \frac{f \left(\right. x \left.\right) - f \left(\right. −10 \left.\right)}{x + 10} & = \underset{x \rightarrow − 10^{+}}{\text{lim}} \frac{- \frac{1}{4} x + \frac{5}{2} - 5}{x + 10} \\ & = \underset{x \rightarrow − 10^{+}}{\text{lim}} \frac{− \left(\right. x + 10 \left.\right)}{4 \left(\right. x + 10 \left.\right)} \\ & = - \frac{1}{4} . \end{aligned}\]

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

Checkpoint 3.8

Find values of \(a\) and \(b\) that make \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. a x + b \text{if} x < 3 \\ x^{2} \text{if} x \geq 3 \end{aligned}\) both continuous and differentiable at \(3 .\)