# inverse function

• Feb 4th 2007, 11:03 PM
gracy
inverse function
1>what is the value of d/dx(f^-1 (x)) when x=2 and f(x)=x^3+x and f^-1(2)=1
(a) 1/13 (b) 1/4 (c) 1/2 (d)7/4

2>Inverse of f(x)= cuberoot((1+x)/2x)
(a)(2x/(1+x))^3
(b) 1/(2x^3-1)
(c)(1/(2x-x))^3
(d)2x/(x^3-1)
• Feb 5th 2007, 03:38 AM
topsquark
Quote:

Originally Posted by gracy
2>Inverse of f(x)= cuberoot((1+x)/2x)
(a)(2x/(1+x))^3
(b) 1/(2x^3-1)
(c)(1/(2x-x))^3
(d)2x/(x^3-1)

Here's how to find an inverse:
Given the function y = f(x), switch the roles of x and y. Thus you have the equation x = f(y). Now solve for y and that gives you your inverse function.

Example:
$\displaystyle f(x) = \sqrt[3]{\frac{1+x}{2x}}$

Set $\displaystyle y = f(x) = \sqrt[3]{\frac{1+x}{2x}}$

Now switch the x and y:
$\displaystyle x = \sqrt[3]{\frac{1+y}{2y}}$

Now solve for y:
$\displaystyle x^3 = \frac{1+y}{2y}$

$\displaystyle 2yx^3 = 1+y$

$\displaystyle 2yx^3 - y = 1$

$\displaystyle y(2x^3 - 1) = 1$

$\displaystyle y = \frac{1}{2x^3 - 1}$

Thus:
$\displaystyle f^{-1}(x) = \frac{1}{2x^3 - 1}$

-Dan