1 Answer

You cannot find f^-1(x) explicitly, but you can find f^-1(3) and (f^-1)'(3) by using the definitions of inverse functions.

Since f'(x) = 5x^4 + 6x^2 + 1 > 0, we see that f(x) is a monotonically increasing function. Thus, f^-1(x) exists for all x.

By definition, if f(a) = b, then f^-1(b) = a. So, the value of f^-1(3) is the solution to the equation:
x^5 + 2x^3 + x - 1 = 3,

which, by inspection, is x = 1. Thus:
f^-1(3) = 1.

To calculate (f^-1)'(3), note that, by definition:
f[f^-1(x)] = x.

By differentiating with the Chain Rule:
(f^-1)'(x) * f'[f^-1(x)] = 1 ==> (f^-1)'(x) = 1/f'[f^-1(x)].

Therefore:
(f^-1)'(3) = 1/f'[f^-1(3)]
= 1/f'(1)
= 1/[5(1)^4 + 6(1)^2 + 1]
= 1/12.

I hope this helps!

12 years ago. Rating: 1
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