Question

For which of the following decreasing functions ƒ does (ƒ-¹)'(10) = −1/8?

A) f(x)=-5x+15
B) f(x)=-2x^3-2x+14
C) f(x)=-x^5-4x+15
D) f(x)=e^(-2x)-x+9

Answer 1

Recall the inverse function theorem; if
`f(x)` is locally invertible around
`x = a`, and
`f(a) = b` and
`a = f^(-1)(b)`, then

`f\left(f^(-1)(x)\right) = x \implies f'\left(f^(-1)(x)\right) \left(f^(-1)\right)'(x) = 1 \\\\ \implies \left(f^(-1)\right)'(x) = \frac1{f'\left(f^(-1)(x)\right)} \\\\ \implies \left(f^(-1)\right)'(b) = \frac1{f'(a)}`

Then the given derivative value tells us that there is some value of
`x=a` for which
`f(a)=10` and
`f'(a) = -8`. Compute the derivative of each function and check if both of these conditions are met.

• If
  `f(x)=-5x+15`, then
  `f'(x)=-5`. (no)
• If
  `f(x)=-2x^3-2x+14`, then
  `f'(x)=-6x^2-2`. Then
  `f'(a) = -6a^2-2=-8 \implies a = \pm 1`. Now,
  `f(-1)=18` doesn't work, but
  `f(1)=10` does. (B is the correct choice)
• If
  `f(x)=-x^5-4x+15`, then
  `f'(x)=-5x^4 - 4`. Then
  `f'(a)=-5a^4-4=-8 \implies a = \pm\sqrt[4]{\frac45}`. But
  `f(a)\\eq10` for either of these values. (no)
• If
  `f(x)=e^(-2x)-x+9`, then
  `f'(x)=-2e^(-2x)-1`. Then
  `f'(a)=-2e^(-2a)-1=-8 \implies a = \frac{\ln(2)-\ln(7)}2`. But
  `f(a)\\eq10`. (no)