Question

F(x)=middle" class="latex-formula">, x$\leq \}$0
find $(f^-^1)'(1)$


I keep getting the wrong answer. The correct answer is$\frac{1}{\sqrt{5} }$

Answer 1

Solve for x when √(x ² - 4) = 1 :

√(x ² - 4) = 1

x ² - 4 = 1

x ² = 5

x = ±√5

We're looking at x ≤ 0, so we take the negative square root, x = -√5.

This means f (-√5) = 1, or in terms of the inverse of f, we have f ⁻¹(1) = -√5.

Now apply the inverse function theorem:

If f(a) = b, then  (f ⁻¹)'(b) = 1 / f '(a).

We have

f(x) = √(x ² - 4)   →   f '(x) = x / √(x ² - 4)

So if a = -√5 and b = 1, we get

(f ⁻¹)'(1) = 1 / f ' (-√5)

(f ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5

The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of f at x = 1.