The function f(x) is an increasing function about which little else is known about other than f(2)=7 and f'(2)=5. Find (f^-1)'(7).

Since f(x) is (strictly) increasing, we know that it is one-to-one and has an inverse f^(-1)(x). Then we can apply the inverse function theorem. Suppose f(a) = b and a = f^(-1)(b). By definition of inverse function, we have

f^(-1)(f(x)) = x

Differentiating with the chain rule gives

(f^(-1))'(f(x)) f'(x) = 1

so that

(f^(-1))'(f(x)) = 1/f'(x)

Let x = a; then

(f^(-1))'(f(a)) = 1/f'(a)

(f^(-1))'(b) = 1/f'(a)

In particular, we take a = 2 and b = 7; then

(f^(-1))'(7) = 1/f'(2) = 1/5

mx+89b(-98)nb

step-by-step explanation: