Hi,
I would like some help with a problem.
Show that $\displaystyle F(x) = \frac{4x^3}{x^2+1}$ has an inverse and find
$\displaystyle f^{-1}´(2)$
How do you invert fractions like this?
//Jones
$\displaystyle f(x) = \frac{4x^3}{x^2+1}$ ... domain is all reals.
$\displaystyle f'(x) = \frac{4x^2(x^2+3)}{(x^2+1)^2} > 0$ for all x, telling us that f(x) is a strictly increasing function.
since f(x) is strictly increasing, it is 1-1 over its domain, and as such has an inverse function.
next, note that $\displaystyle f(1) = 2$
that means that $\displaystyle f^{-1}(2) = 1$
Note, that this problem did NOT ask you to actually find the inverse function- and skeeter did not. However, since you asked:
Start by witing the function as
$\displaystyle f(x) = y= \frac{4x^3}{x^2+1}$ and swapping x and y:
$\displaystyle x= \frac{4y^3}{y^2+ 1}$
Now solve for y:
$\displaystyle x(y^2+ 1)= 4y^3$
$\displaystyle 4y^3- xy^2+ x= 0$
Solve that cubic equation for y. That, of course, is the hard part and probably why the problem did not require you to find the inverse!