Suppose f is a polynomial function that has an inverse f^-1.

Question

Find (f^-1)'(2) if f(x)=x^3 + 2x - 1?

Answer 1

The polynomial function is f(x) = x³ + 2x - 1 and it has inverse f^-1(x) (itself).

So, the function f(x) is one-to-one ⇒ f(x) contains (a, b) and f^-1(x) contains (b, a) ⇒ f(a) = b then f^-1(b) = a.

To find (f ^-1)^'(2), first adjusted f(x) = 2 at which value of x.

The value of x = 1 then f(x) = (1)³ + 2(1) - 1 = 2.

f(1) = 2 then f^-1(2) = 1.

The function f(x) is differetiable and has an inverse function, so can apply theorem g^'(x) = 1 / [ f^'(g(x)) ], f^'(g(x)) ≠ 0.

Let g(x) = (f ^-1)(x) then (f ^-1)^'(x) = 1 / [ f^'(f ^-1(x)) ].

(f ^-1)^'(2) = 1 / [ f^'(f ^-1(2)) ] = 1 / [ f^'(1) ].

Moreover, using f^'(x) = 3x² + 2, conclude that

(f ^-1)^'(2) = 1 / [ f^'(1) ] = 1 / [ 3(1)² + 2 ] = 1 / [ 3(1)² + 2 ] = 1/5.

The value of (f ^-1)^'(2) = 1/5.