Wemazi

Topic 3.3

Check when the inverse rule applies

Builds on using inverse information from data. 7 minutes.

One output, two inputs

The curve is \(f(x) = x^2 - 6x\), and its lowest point is the vertex \((3, -9)\). The dashed line at height \(-5\) meets the curve at two orange points, so the inputs 1 and 5 both have the output \(-5\).

An inverse can return only one input, so it uses one branch, a piece of the curve where f only rises or only falls. The buttons choose the branch to the right or left of the vertex and draw it bold. Its reflection across the faint diagonal is the graph of the inverse.

On the right branch, the matching input is 5, where the tangent has slope 4. This tangent reflects into the mirror tangent of the inverse, which has slope \(\frac14\).

On the left branch, the input is 1, and the tangent and mirror tangent have slopes \(-4\) and \(-\frac14\), so the branch decides the sign.

Choose a branch and test the tangent

The line at height \(b\) meets the curve at two points, and each is a possible matching input \(a\), an input with \(f(a) = b\).

The inverse uses the branch that holds \(a\), where each height has one input. The slope of the mirror tangent is the reciprocal of the slope of \(f\) at \(a\).

Lower the line to the turning point. The tangent of f turns flat, and its mirror tangent turns vertical. A vertical line has no slope, so the inverse has no derivative there.

Notation

  1. Choose a branch of \(f\) that contains the matching input \(a\), where \(f(a) = b\). On a branch, \(f\) only rises or only falls, so each output has one input.

  2. If \(f'\) has one sign on an interval, that interval is a branch. The curve is green where \(f' > 0\) and \(f\) rises, and red where \(f' < 0\) and \(f\) falls.

    \[f' > 0 \ \text{or}\ f' < 0 \ \text{on the interval}\]

  3. The function \(f\) must be differentiable at \(a\), which means it has a tangent there with slope \(f'(a)\).

  4. The derivative \(f'(a)\) must not be 0, because the rule divides by it.

    \[(f^{-1})'(b) = \frac{1}{\cy{f'(a)}}, \qquad f'(a) \neq 0\]

  5. If \(f'(a) = 0\), the derivative of the inverse at \(b\) does not exist.

When it fails

At height \(-9\), the line touches the curve only at the vertex \((3, -9)\), where the slope of \(f\) is 0.

The mirror tangent is vertical, and a vertical line has no slope. The formula would need \(\frac10\), so \((f^{-1})'(-9)\) does not exist.

Example 1: Test the rule, then use it

Let \(f(x) = x + \ln x\) for \(x > 0\). Find \((f^{-1})'(e + 1)\), and show that the rule applies.

  1. The derivative is positive for \(x > 0\), so \(f\) only rises and the whole curve is one branch, drawn green.

    \[f'(x) = 1 + \frac{1}{x} > 0\]

  2. Because \(\ln e = 1\), the matching input is \(e\).

    \[f(e) = e + \ln e = \cy{e + 1}\]

  3. The function is differentiable at \(e\), and its slope there is not 0.

    \[f'(e) = 1 + \frac{1}{e} \neq 0\]

  4. All three tests pass. Take the reciprocal of the slope at \(e\).

    \[(f^{-1})'(e + 1) = \frac{1}{1 + \frac1e} = \cy{\frac{e}{e + 1}}\]

Example 2: A flat tangent

Let \(f(x) = (x - 2)^5 + 3\). Find \((f^{-1})'(3)\).

  1. A fifth power only rises, so the whole curve is one branch, drawn green. The matching input is 2.

    \[f(2) = 0 + 3 = \cy{3}\]

  2. At 2, the slope of \(f\) is 0, so the tangent is flat.

    \[f'(x) = 5(x - 2)^4, \qquad f'(2) = \cy{0}\]

  3. A tempting wrong move: The slope of \(f\) at 2 is 0, so the slope of the inverse is also 0.

    \[(f^{-1})'(3) = \cr{0}\]

    A reflection across the diagonal turns the flat tangent at \((2, 3)\) into a vertical line through \((3, 2)\), not into the flat red line.

  4. The rule divides by \(f'(2) = 0\), which is not allowed. The mirror tangent is vertical, so \((f^{-1})'(3)\) does not exist.

Practice

Let \(f(x) = x^2 - 8x\) for \(x \le 4\). Find \((f^{-1})'(-12)\).

Solve f(a) = −12, and keep the solution on the bold branch x ≤ 4, where the curve falls.

Solution

  1. Solve \(f(a) = -12\). Two inputs have this output, and only one lies on the branch \(x \le 4\).

    \[\begin{aligned} a^2 - 8a + 12 &= 0 \\ (a - 2)(a - 6) &= 0 \\ a &= \cy{2} \end{aligned}\]

  2. Find the slope of \(f\) at 2. It is not 0, and it is negative because \(f\) falls on this branch.

    \[f'(x) = 2x - 8, \qquad f'(2) = -4\]

  3. Take the reciprocal.

    \[(f^{-1})'(-12) = \frac{1}{f'(2)} = \cy{-\frac14}\]

Questions

The graph shows \(f(x) = x^3 - 3x\). Its tangent is horizontal at \((-1, 2)\) and at \((1, -2)\).

  1. Use only the part of the curve with \(x \ge 1\). At which output \(b\) does its inverse have no derivative?

    • \(b = -2\)
    • \(b = 2\)
    • \(b = 1\)
    • \(b = 0\)
    Answer

    \(b = -2\)

  2. On which interval is \(f\) one-to-one, so that it has an inverse there?

    • \([1.5,\,3]\)
    • \([-2,\,0]\)
    • \([0,\,2]\)
    • \([-2,\,2]\)
    Answer

    \([1.5,\,3]\)

  3. Let \(g(x) = x^3 - 12x\) for \(x \ge 2\). What is \((g^{-1})'(16)\)?

    • \(\tfrac{1}{36}\)
    • \(\tfrac{1}{756}\)
    • The derivative does not exist
    • \(\tfrac14\)
    Answer

    \(\tfrac{1}{36}\)

Summary

Before you use the inverse rule, you check that \(a\) lies on a branch, that \(f\) is differentiable at \(a\), and that \(f'(a) \neq 0\).

The branch gives each output one input, and the slope of the inverse is the reciprocal of the slope of f.

If \(f'(a) = 0\), the flat tangent of \(f\) reflects into a vertical tangent, so you write that the derivative of the inverse does not exist.

This is one of four skills in topic 3.3 of Which & Why: AP Calculus.