f(x) = x^3 + 3sin x + 2cos x | Find (ƒ^(-1))'(2)

Question

ƒ(x) = x³ + 3sin x + 2cos x

Find (ƒ^-1)'(2)

Answer 1

First, find out what x value satisfies the equation f(x) = 2; we'll need this to determine the inverse value.

2 = x³ + 3sinx + 2cosx

Solving this straight up would be rather difficult, but we can use a guess and check to find the solution. Since trig is involved, I'd start with x=0: 2 = 0³ + 3sin(0) + 2cos(0) = 0 + 3•0 + 2•1 = 2. Since this works, we'll say that f(0) = 2 is the value of interest.

Now, we'll need f'(x) since [f^-1(x₁)]' = 1/[f'(x₂)] where f(x₂) = x₁. f'(x) = 3x² + 3cosx - 2sinx, then f'(0) = 3. Plugging this into the above formula, we get [f^-1(2)]' = ¹/₃.

Thus the inverse derivative at x=2 is ¹/₃.

Answer 2