Calculus Homework

Question

Find (f^-1)'(a).

f(x)=5x^3+4x^2+6x+8, a=8

(f^-1)'(a)=

Answer 1

Question

Given f(x) = y = 5x³ + 4x²+ 6x +8, find (f^-1)'(a) at a = 8.

To find (f^-1)' switch x and y

x = 5y³ + 4y²+ 6y +8

then differentiate with respect to x, and solve for dy/dx

dx/dx = 1 = (15y² + 8y +6)(dy/dx) which means that

dy/dx = 1/(15y² + 8y + 6)

This is (f^-1)', the derivative of the inverse function, but its written in terms of y.

To find (f^-1)'(8) we need to ask, for the inverse function f^-1(x): What does f^-1(x) equal when x = 8?

f^-1(8) = ?

To answer this we ask, for the original function f(x): For what value of x does f(x) = 8?

f(?) = 8

Looking at f(x) = 5x³ + 4x²+ 6x + 8, we can readily see that f(x) = 8 when x = 0, or f(0) = 8

If f(0) = 8, then f^-1(8) = 0

So to determine (f^-1)'(8) we evaluate the derivative dy/dx = 1/(15y² + 8y + 6) at y = 0.

(f^-1)'(8) = 1/(15(0)² + 8(0) + 6) = 1/6

Answer

(f^-1)'(8) = 1/6

Answer 2

In order to find (f^-1)^'(8) we first need to find an x-value such that f(x) = 8. By inspection, f(0) = 8.

Now, we need f^'(x) which we can get by differentiating using power rule:

f^'(x) = 15x² + 8x + 6

f^'(0) = 6.

So, (f^-1)^'(8) = 1/6.