Question

F(x) = x³ + 2x - 1, a = -4
(f⁻¹)′( - 4) = ?

Answer 1

Recall the inverse function theorem. If
`f` is invertible and differentiable in all the right places, then

`f^(-1)\bigg(f(x)\bigg) = x \\\\ ~~~~~~~~ \implies \left(f^(-1)\right)'\bigg(f(x)\bigg) f'(x) = 1 \\\\ ~~~~~~~~ \implies \left(f^(-1)\right)'\bigg(f(x)\bigg) = \frac1{f'(x)} \\\\ ~~~~~~~~ \implies \left(f^(-1)\right)'(x) = \frac1{f'\left(f^(-1)(x)\right)} \\\\ ~~~~~~~~ \implies \left(f^(-1)\right)'(-4) = \frac1{f'\left(f^(-1)(-4)\right)}`

Solve for
`x` such that
`f(x)=-4`.

`x^3 + 2x - 1 = -4 \\\\ x^3 + 2x + 3 = 0 \\\\ (x + 1) (x^2 - x + 3) = 0`

`x^2-x+3=0` has non-real solutions, so we're left with

`x+1 = 0 \implies x = -1`

Then we have

`f(-1) = -4 \implies f^(-1)(-4) = -1 \implies \left(f^(-1)\right)' (-4) = \frac1{f'(-1)} = \boxed{\frac15}`

since

`f(x) = x^3 + 2x - 1 \implies f'(x) = 3x^2 + 2 \implies f'(-1)=5`