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Coordinated Calculus

Nathan Wakefield, Christine Kelley, Marla Williams, Michelle Haver, Lawrence Seminario-Romero, Robert Huben, Aurora Marks, Stephanie Prahl, Based upon Active Calculus by Matthew Boelkins

Section 1.8 Differentiability

In this section we aim to determine whether or not the function has a derivative \(f'(a)\) at \(x = a\text{.}\)

Subsection 1.8.1 Being Differentiable at a Point

We recall that a function \(f\) is said to be differentiable at \(x = a\) if \(f'(a)\) exists. Moreover, for \(f'(a)\) to exist, we know that the graph of \(y = f(x)\) must have a tangent line at the point \((a,f(a))\text{,}\) since the value of \(f'(a)\) is precisely the slope of this line. Observe that in order to ask if \(f\) has a tangent line at \((a,f(a))\text{,}\) it is necessary for \(f\) to be continuous at \(x = a\text{:}\) if \(f\) fails to have a limit at \(x = a\text{,}\) if \(f(a)\) is not defined, or if \(f(a)\) does not equal the value of \(\lim_{x \to a} f(x)\text{,}\) then it doesn’t make sense to talk about a tangent line to the curve at this point.
Indeed, it can be proved formally that if a function \(f\) is differentiable at \(x = a\text{,}\) then it must be continuous at \(x = a\text{.}\) Stated differently, if \(f\) is not continuous at \(x = a\text{,}\) then it is automatically the case that \(f\) is not differentiable there. For example, in Figure 1.96 below, both \(f\) and \(g\) fail to be differentiable at \(x = 1\) because neither function is continuous at \(x = 1\text{.}\)
Figure 1.96. Functions \(f\) and \(g\) that are not continuous (and hence not differentiable) at \(x=1\text{,}\) and a function \(h\) that is both continuous and differentiable at \(x=1\text{.}\)
A natural question to ask at this point is “is there a difference between continuity and differentiability?” In other words, can a function fail to be differentiable at a point where the function is continuous? To answer these questions, we consider a certain function \(f\text{,}\) where the graph of \(y=f(x)\) is displayed below in Figure 1.97. We notice that \(f\) has a sharp corner at the point \((1,1)\text{,}\) and further observe that \(f\) is continuous at \(x=1\) since \(\lim_{x\to1} f(x)=1=f(1)\text{.}\)
Figure 1.97. A function \(f\) that is continuous at \(x = 1\) but not differentiable at \(x = 1\text{;}\) at right, we zoom in on the point \((1,1)\) in a magnified version of the box shown in the left-hand plot.
However, the function \(f\) in Figure 1.97 is not differentiable at \(x = 1\) because \(f'(1)\) fails to exist. One way to see this is to observe that \(f'(x) = -1\) for every value of \(x\) that is less than 1, while \(f'(x) = +1\) for every value of \(x\) that is greater than 1. That makes it seem that either \(+1\) or \(-1\) would be equally good candidates for the value of the derivative at \(x = 1\text{.}\) Alternatively, we could use the limit definition of the derivative to attempt to compute \(f'(1)\text{,}\) and discover that the derivative does not exist. Finally, we can see visually in Figure 1.97 that this function does not have a tangent line at \(x=1\text{.}\) Regardless of how closely we examine the function by zooming in on \((1,1)\) on the graph of \(y=f(x)\text{,}\) it will always look like a “V” and never like a single line, which tells us there is no possibility for a tangent line there.

Subsection 1.8.2 Local Linearity

If a function does have a tangent line at a given point, then the function and the tangent line should appear essentially indistinguishable when we zoom in on the point of tangency.
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For instance, see http://gvsu.edu/s/6J for an applet (due to David Austin, GVSU) where zooming in shows the increasing similarity between the tangent line and the curve.
Conversely, if we zoom in on a point and the function looks like a single straight line, then the function should have a tangent line there and thus be differentiable at that point. Therefore, we say that a function that is differentiable at \(x = a\) is locally linear. An example of this can be seen in Figure 1.98 below, with the graph of a function \(y=f(x)\) on the left and a magnified version with the tangent line to \(f\) at \(a\) on the right.
Figure 1.98. A function \(y = f(x)\) and its tangent line at the point \((a,f(a))\) is shown in the figure. Notice how the function resembles a line as we zoom in on \(x=a\text{.}\)

Example 1.99.

Let \(g\) be the function given by the rule \(g(x) = |x|\text{.}\)
  1. Visualize (or sketch) the graph of \(y=g(x)\text{,}\) and use it to explain why \(g\) is differentiable at every nonzero value of \(x\text{.}\)
  2. Use the limit definition of the derivative to show that \(g'(0) = \lim_{h \to 0} \frac{|h|}{h}\text{.}\) Then use left- and right- handed limits to explain why this limit does not exist. Conclude that \(g\) is not differentiable at \(x=0\text{.}\)
Hint.
  1. What type of function is \(g\) for all \(x \lt 0\text{?}\) For all \(x > 0\text{?}\)
  2. Recall that \(g'(0) = \lim_{h \to 0} \frac{g(0 + h) - g(0)}{h}\text{.}\) What is the value of \(|h|\) when \(h \lt 0\text{?}\)
Answer.
  1. \(g\) is piecewise linear.
  2. \begin{equation*} \lim_{h \to 0^-} \frac{|h|}{h}=-1, \ \text{and} \ \lim_{h\to0^+}\frac{|h|}h=1\text{.} \end{equation*}
Solution.
  1. We know that \(g(x) = |x|\) is given by the formula \(g(x) = -x\) when \(x \lt 0\) and by \(g(x) = x\) when \(x \ge 0\text{.}\) Each of these pieces of \(g\) is a straight line, so at every point other than the point where they meet, the function \(g\) has a well-defined slope and thus is differentiable whenever \(x\neq0\text{.}\) The graph of \(y=g(x)\) is a translate of the earlier graph of \(y=f(x)\) from Figure 1.97.
  2. Observe that
    \begin{align*} g'(0) =\mathstrut \amp \lim_{h \to 0} \frac{g(0+h)-g(0)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|0+h|-|0|}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|h|}{h} \end{align*}
    To evaluate this limit, note that \(|h| = h\) whenever \(h > 0\text{,}\) and thus
    \begin{equation*} \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\text{.} \end{equation*}
    In contrast, we have \(|h| = -h\) whenever \(h \lt 0\text{,}\) and thus
    \begin{equation*} \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\text{.} \end{equation*}
    Since the right- and left-hand limits are not equal, it follows that
    \begin{equation*} g'(0) = \lim_{h \to 0} \frac{|h|}{h} \end{equation*}
    does not exist. Therefore, the function \(g(x)=|x|\) is not differentiable at \(x=0\text{.}\)

Subsection 1.8.3 Vertical Tangent Lines

Another example of when a function can fail to be differentiable at a point \(x=a\) is if the function has a vertical tangent at the point. In other words, when \(f\) is continuous at \(x=a\) and \(\lim_{x\to a}|f'(x)|=\infty\text{.}\) This means the tangent lines become very steep as we move closer to \(x=a\text{.}\)

Example 1.100.

In this example, let \(f(x)=\sqrt{x}\text{.}\)
In Figure 1.101 below, we have the graph of \(y=f(x)\) along with a progression of tangent lines at points approaching \((0,0)\) on the graph. As we approach \(x=0\text{,}\) we see that the tangent lines drawn become steeper and steeper, ultimately leading to a vertical tangent line at \(x=0\text{.}\)
Figure 1.101. As we move closer to \(x=0\text{,}\) the tangent lines to the graph of \(y=f(x)\) become steeper and steeper. Notice that the tangent line closest to \(x=0\) is nearly vertical.
We can also show this by calculating the limit of the derivative close to \(x=0\text{:}\)
\begin{equation*} \lim_{x\to 0}f'(x)=\lim_{x\to 0}\frac{1}{2\sqrt{x}}=\infty. \end{equation*}
Therefore, \(f(x)\) is not differentiable at \(x=0 \text{.}\)

Subsection 1.8.4 Links Between Continuity, Differentiability, and Limits

To summarize the preceding discussion of differentiability, we make several important observations.
  • If \(f\) is differentiable at \(x = a\text{,}\) then \(f\) is continuous at \(x = a\text{.}\) Equivalently, if\(f\) fails to be continuous at \(x = a\text{,}\) then \(f\) will not be differentiable at \(x = a\text{.}\)
  • A function can be continuous at a point without being differentiable there. In particular, a function \(f\) is not differentiable at \(x = a\) if the graph has a sharp corner (or cusp) at the point \((a,f(a))\text{.}\)
  • If \(f\) is differentiable at \(x = a\text{,}\) then \(f\) is locally linear at \(x = a\text{.}\) In other words, a differentiable function looks linear when viewed up close because it resembles its tangent line at any given point of differentiability.

Example 1.102.

In this example, let \(f\) be the function whose graph is given below in Figure 1.103.
Figure 1.103. The graph of \(y = f(x)\) for Example 1.102.
  1. State all values of \(a\) for which \(f\) is not continuous at \(x = a\text{.}\) For each, provide a reason for your conclusion.
  2. State all values of \(a\) for which \(f\) is not differentiable at \(x = a\text{.}\) For each, provide a reason for your conclusion.
  3. State all values of \(a\) for which \(f\) is not differentiable, but is continuous at \(x = a\text{.}\) Think about why this is the case.
Hint.
  1. You might start by looking for places you would lift your pencil when drawing the graph, identifying points where \(f\) is not continuous. Calculating limits may also be useful.
  2. You might start by identifying points where \(f\) is not continuous.
  3. Thinking back to Example 1.99 may be useful.
Answer.
  1. At \(a = -2 \text{,}\) \(\lim_{x\to-2}f(x)\) does not exist; at \(a=-1\text{,}\) \(\lim_{x\to-1}f(x)\neq f(-1)\text{;}\) at \(a=2\text{,}\) \(\lim_{x\to2}f(x)\) does not exist; at \(a=3\text{,}\) \(f(3)\) is undefined.
  2. \(a = -2, -1, 2, 3\text{,}\) because \(f\) is not continuous at these points; \(a=-3,1\text{,}\) because \(f\) does not have a tangent line at these points.
  3. \(a = -3,1\text{.}\)
Solution.
  1. \(f\) is not continuous at \(a = -2, 2\) because at each of these points \(\lim_{x\to a}f(x)\) does not exist. \(f\) is not continuous at \(a = -3\) because \(\lim_{x\to -1}f(x)=-3.5\text{,}\) but \(f(-1)=1\text{.}\) \(f\) is not continuous at \(a=3\) because \(f(3)\) is not defined.
  2. \(f\) is not differentiable at \(a = -2, -1, 2, 3\) because at each of these points \(f\) is not continuous. In addition, \(f\) is not differentiable at \(a = -3\) and \(a = 1\) because the graph of \(f\) has a corner point (or cusp) at each of these values.
  3. The only two points where \(f\) is continuous but not differentiable are \(a=-3,1\text{.}\) This is because of the corner point (or cusp). These points fit the criteria for continuity, but there is no discernible tangent line.

Example 1.104.

True or false: if a function \(p\) is differentiable at \(x = b\text{,}\) then \(\lim_{x \to b} p(x)\) must exist. Write at least one sentence to justify your choice.
Hint.
What does being differentiable at a point tell you about continuity there?
Answer.
True.
Solution.
We know that a function \(f\) is continuous whenever it is differentiable, and that one characteristic of \(f\) being continuous at \(x=a\) is that \(\lim_{x\to a}f(x)\) exists. Therefore the statement “if a function \(p\) is differentiable at \(x = b\text{,}\) then \(\lim_{x \to b} p(x)\) must exist” is true.

Subsection 1.8.5 Summary

  • A function \(f\) is differentiable at \(x = a\) whenever \(f'(a)\) exists, which means that \(f\) has a tangent line at \((a,f(a))\) and thus \(f\) is locally linear at \(x = a\text{.}\) Informally, this means that the function looks like a line when viewed up close at \((a,f(a))\) and that there is not a corner point or cusp at \((a,f(a))\text{.}\)
  • Differentiability is a stronger condition than continuity, which is a stronger condition than having a limit. In particular, if \(f\) is differentiable at \(x = a\text{,}\) then \(f\) is also continuous at \(x = a\text{,}\) and if \(f\) is continuous at \(x = a\text{,}\) then \(f\) has a limit at \(x = a\text{.}\)

Exercises 1.8.6 Exercises

1. Continuity and differentiability of a graph.

Consider the function graphed below.
At what \(x\)-values does the function appear to not be continuous? \(x =\)
At what \(x\)-values does the function appear to not be differentiable? \(x =\)
(Enter none if there are no \(x\)-values that apply; enter \(x\)-values as a comma-separated list, e.g., 1,3,5.)

2. Continuity and differentiability of a graph.

Consider the graph of the function \(y = p(x)\) that is provided in Figure 1.105. Assume that each portion of the graph of \(p\) is a straight line, as pictured.
Figure 1.105. At left, the piecewise linear function \(y = p(x)\text{.}\) At right, axes for plotting \(y = p'(x)\text{.}\)
  1. State all values of \(a\) for which \(\lim_{x \to a} p(x)\) does not exist.
  2. State all values of \(a\) for which \(p\) is not continuous at \(a\text{.}\)
  3. State all values of \(a\) for which \(p\) is not differentiable at \(x = a\text{.}\)
  4. On the axes provided in Figure 1.105, sketch an accurate graph of \(y = p'(x)\text{.}\)

3. Examples of functions.

For each of the following prompts, give an example of a function that satisfies the stated criteria; a formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why.
  1. A function \(f\) that is continuous at \(a = 2\) but not differentiable at \(a = 2\text{.}\)
  2. A function \(g\) that is differentiable at \(a = 3\) but does not have a limit at \(a=3\text{.}\)
  3. A function \(h\) that has a limit at \(a = -2\text{,}\) is defined at \(a = -2\text{,}\) but is not continuous at \(a = -2\text{.}\)
  4. A function \(p\) that satisfies all of the following:
    • \(p(-1) = 3\) and \(\lim_{x \to -1} p(x) = 2\)
    • \(p(0) = 1\) and \(p'(0) = 0\)
    • \(\lim_{x \to 1} p(x) = p(1)\) and \(p'(1)\) does not exist

4. Estimating the derivative at a point.

Consider the function \(g(x) = \sqrt{|x|}\text{.}\)
  1. Use a graph to explain visually why \(g\) is not differentiable at \(x = 0\text{.}\)
  2. Use the limit definition of the derivative to show that
    \begin{equation*} g'(0) = \lim_{h \to 0} \frac{\sqrt{|h|}}{h}\text{.} \end{equation*}
  3. Investigate the value of \(g'(0)\) by estimating the limit in (b) using small positive and negative values of \(h\text{.}\) For instance, you might compute \(\frac{\sqrt{|-0.01|}}{0.01}\text{.}\) Be sure to use several different values of \(h\) (both positive and negative), including ones closer to 0 than 0.01. What do your results tell you about \(g'(0)\text{?}\)
  4. Use your graph in (a) to sketch an approximate graph of \(y = g'(x)\text{.}\)
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