Recognize an unbounded slope
A curve that climbs straight up
The graph of \(f(x) = \sqrt[3]{x-1}\) is continuous and climbs steeply through the orange dot \((1, 0)\). The orange line joins the dot to a point on the right, and the chalk line joins it to a point on the left.
Press "Distance 0.1", then "Distance 0.001", then "Distance 0.000001". The slope from the right is about 4.64, then 100, then 10,000. The slope from the left is the same each time, and it is positive because the graph climbs.
These slopes grow without bound, which means they become larger than any number you name. They do not approach a number, so \(f'(1)\) does not exist.
Both lines turn toward the vertical line through the dot, which is called a vertical tangent and has no slope.
Large slopes with the same sign
The graph is continuous at P, and its two arms are the parts on each side of P. Move the two points toward P, one on each arm.
The orange line to the right arm and the chalk line to the left arm both become steeper. Their slopes have the same sign and grow without bound, so the lines turn toward one vertical line.
That vertical line is the tangent at P. The derivative at P does not exist, because a vertical line has no slope.
Notation
A graph has a vertical tangent at \(a\) if \(f\) is continuous at \(a\). The slopes of the lines through \((a, f(a))\) must also grow without bound, with the same sign on both sides.
\[\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a} = \cy{\pm\infty}\]If the limit is \(+\infty\), the graph climbs through the point. If it is \(-\infty\), the graph falls.
Because \(\infty\) is not a number, the limit is not a number, so \(f'(a)\) does not exist.
A large slope alone is not enough. If the slopes approach 10,000, then \(f'(a) = 10{,}000\), but here they grow without bound.
When it fails
After the flip, the slopes from the right are large and positive, but those from the left are large and negative. The graph falls into P and climbs out of it, so both arms lie above P.
The two arms meet in a sharp point called a cusp. It has no vertical tangent, because the signs differ, and the derivative does not exist.
Example 1: a falling vertical tangent
Let \(f(x) = 5 - \sqrt[3]{x-4}\). Does the graph have a vertical tangent at \(x = 4\)?
The function is continuous at 4, and \(f(4) = 5\). Write the slope of the line from \((4, 5)\) to the point at \(x = 4 + h\).
\[\begin{aligned} \dfrac{f(4+h)-f(4)}{h} &= \dfrac{-\sqrt[3]{h}}{h} \\ &= -\dfrac{1}{h^{2/3}} \end{aligned}\]The power \(h^{2/3}\) is the square of \(\sqrt[3]{h}\), so it is positive for both signs of \(h\) and approaches 0. The slope is therefore negative and grows without bound on both sides. At \(h = \pm 0.001\) it is −100.
The signs agree, so the graph has a vertical tangent and falls through the dot. The derivative does not exist.
\[\cy{\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4} = -\infty}\]
Example 2: a sharp point
Let \(g(x) = (x+3)^{2/3}\). Does the graph have a vertical tangent at \(x = -3\)?
Write the slope of the line with \(h = x + 3\). Since \(g(-3) = 0\), the slope is \(h^{2/3}/h\), which is \(1/h^{1/3}\).
\[\dfrac{g(-3+h)-g(-3)}{h} = \dfrac{1}{h^{1/3}}\]A tempting wrong move: The slope grows without bound, so the graph has a vertical tangent.
\[\cr{\left|\dfrac{1}{h^{1/3}}\right| \to \infty}\]The slope is 10 for \(h = 0.001\) and −10 for \(h = -0.001\), so the signs differ. Both arms lie above the dot, so no part of the graph runs along the red line below it.
Because \(h^{1/3}\) has the sign of \(h\), the slope tends to \(+\infty\) from the right and to \(-\infty\) from the left. The graph has a cusp, not a vertical tangent, and the derivative does not exist.
\[\begin{aligned} \cy{\lim_{h\to 0^+}\dfrac{1}{h^{1/3}}} &= \cy{+\infty} \\ \cy{\lim_{h\to 0^-}\dfrac{1}{h^{1/3}}} &= \cy{-\infty} \end{aligned}\]
Practice
Let \(f(x) = \sqrt[5]{x-3}\). Does the graph have a vertical tangent at \(x = 3\)?
Write the slope as a power of \(h\). Then find its sign for \(h > 0\) and for \(h < 0\).
Solution
Here \(f(3) = 0\). The slope of the line is a power of \(h\).
\[\dfrac{h^{1/5}}{h} = \dfrac{1}{h^{4/5}}\]The power \(h^{4/5}\) is the fourth power of \(\sqrt[5]{h}\), so it is positive for both signs of \(h\) and approaches 0. The slope is therefore positive and grows without bound on both sides. The graph has a vertical tangent, and \(f'(3)\) does not exist.
\[\cy{\lim_{x\to 3}\dfrac{f(x)-f(3)}{x-3} = +\infty}\]
Questions
Questions 1 and 2 use the graph of the cube root through \((1, 0)\). Question 3 needs a calculation.
The orange point moves toward the orange dot from the right. What happens to the slope of the line through them?
- It grows without bound.
- It approaches 1.
- It approaches 0.
- It changes sign again and again.
Answer
It grows without bound.
The left arm is now flipped, so the graph has a sharp point. Does it have a vertical tangent at \(x = 1\)?
- No. The slopes from the right are large and positive, and from the left they are large and negative.
- Yes. The slopes grow without bound.
- Yes. The graph is continuous at \(x = 1\).
Answer
No. The slopes from the right are large and positive, and from the left they are large and negative.
Let \(f(x) = 4\sqrt[3]{x-2}\). The line from \((2, 0)\) to the point at \(x = 2.001\) has slope 400. What does this suggest about \(f'(2)\)?
- The slopes grow without bound as the gap shrinks, so \(f'(2)\) does not exist.
- \(f'(2) = 400\).
- \(f'(2) = 4\).
- \(f'(2) = 0.4\).
Answer
The slopes grow without bound as the gap shrinks, so \(f'(2)\) does not exist.
Summary
You write the slope of the line as a power of h and find its sign on each side.
A vertical tangent occurs when the slope grows without bound with the same sign on both sides, and a cusp occurs when the signs differ.
This test applies to roots and fractional powers, where a graph can be continuous and still have no derivative.
This is one of five skills in topic 2.4 of Which & Why: AP Calculus.