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

$f^{-1}\bigg(f(x)\bigg) = x \\\\ ~~~~~~~~ \implies \left(f^{-1}\right)'\bigg(f(x)\bigg) f'(x) = 1 \\\\ ~~~~~~~~ \implies \left(f^{-1}\right)'\bigg(f(x)\bigg) = \dfrac1{f'(x)} \\\\ ~~~~~~~~ \implies \left(f^{-1}\right)'(x) = \dfrac1{f'\left(f^{-1}(x)\right)} \\\\ ~~~~~~~~ \implies \left(f^{-1}\right)'(-4) = \dfrac1{f'\left(f^{-1}(-4)\right)}$

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

$f(-1) = -4 \implies f^{-1}(-4) = -1 \implies \left(f^{-1}\right)' (-4) = \dfrac1{f'(-1)} = \boxed{\dfrac15}$

since

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