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

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- Dec 30th 2009, 02:05 AMJonesThe inverse of a function
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 - Dec 30th 2009, 05:28 AMskeeter
$\displaystyle f(x) = \frac{4x^3}{x^2+1}$ ... 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 $\displaystyle f^{-1}(2) = 1$ - Dec 30th 2009, 06:03 AMHallsofIvy
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! - Dec 30th 2009, 06:48 AMJones
Oh dang, i thought you had to solve the darn thing