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.
