Find the inverse function, if exists..

**TWiX** · April 8th 2009, 06:02 AM

hello
am stopped at this problem -_-

let $f(x)=2x^2-8x$ where x>2
Find a formula for the inverse function of it, if exists.

My solution:
f`(x)=4x-..... so f is increas..so its 1-1funct..so inverse is exist.

its easy to prove that the inverse function is exist..=)

But the problem is finding the inverse function
ok let $y=2x^2-8x$ ..
how to find the x in terms of y !!
i tried to get the natural logarthmic of both sides but it failed :S

any help?!

**Danny** · April 8th 2009, 06:10 AM

Originally Posted by TWiX

hello
am stopped at this problem -_-

let $f(x)=2x^2-8x$ where x>2
Find a formula for the inverse function of it, if exists.

My solution:
f`(x)=4x-..... so f is increas..so its 1-1funct..so inverse is exist.

its easy to prove that the inverse function is exist..=)

But the problem is finding the inverse function
ok let $y=2x^2-8x$ ..
how to find the x in terms of y !!
i tried to get the natural logarthmic of both sides but it failed :S

any help?!

Switch $x\; \text{and}\; y$ , complete the square in y and then solve for y.

**thelostchild** · April 8th 2009, 06:11 AM

complete the square on the right hand side to give
$y=2x^2-8x$

$\frac{y}{2}=x^2-4x$

$\frac{y}{2}=(x-2)^2-4$

I assume you can take it from there!

**Prove It** · April 8th 2009, 06:17 AM

Originally Posted by TWiX

hello
am stopped at this problem -_-

let $f(x)=2x^2-8x$ where x>2
Find a formula for the inverse function of it, if exists.

My solution:
f`(x)=4x-..... so f is increas..so its 1-1funct..so inverse is exist.

its easy to prove that the inverse function is exist..=)

But the problem is finding the inverse function
ok let $y=2x^2-8x$ ..
how to find the x in terms of y !!
i tried to get the natural logarthmic of both sides but it failed :S

any help?!

$f(x)$ IS one to one, but only because the domain has been restricted to all x-values greater than the turning point, not for the reason you stated.

The domain of f is $x > 2$ , so the range is $y > 0$ , since the function is increasing when $x>2$ ...

Note that for an inverse function, the domain and range swap, which implies that the x and y values swap.

So $f^{-1}(x): x = 2y^2 - 8y$

$\frac{1}{2}x = y^2 - 4y$

$\frac{1}{2}x + \left(-\frac{4}{2}\right)^2 = y^2 - 4y + \left(-\frac{4}{2}\right)^2$

$\frac{1}{2}x + 4 = (y - 2)^2$

Can you solve for y now? Remember that the domain of $f^{-1}$ is $x>0$ since $\textrm{dom}f^{-1} = \textrm{ran}f$ .

**TWiX** · April 8th 2009, 06:52 AM

Get it now ..
Need to solve the equation for x without using the complete of the square ??!

**stapel** · April 8th 2009, 07:19 AM

Originally Posted by TWiX

let $f(x)=2x^2-8x$ where x>2

Find a formula for the inverse function of it, if exists.

Use the standard process:

Rename "f" as "y": y = 2x^2 - 8x

Solve this for "x=". In this case, use the Quadratic Formula:

. . . . . $2x^2\, -\, 8x\, -\, y\, =\, 0$

. . . . . $x\, =\, \frac{-(-8)\, \pm\, \sqrt{(-8)^2\, -\, 4(2)(-y)}}{2(2)}\, =\, \frac{8\, \pm\, \,\sqrt{64\, +\, 8y}}{4}$

...and so forth. In this case, you were given the restriction of x > 2, so you know you need only the positive square root:

. . . . . $x\, =\, 2\, +\, \frac{\sqrt{16\, +\, 2y}}{2}$

Then continue with the usual steps: Swap "x" and "y", and rename the new "y" as f-inverse.

**TWiX** · April 8th 2009, 08:43 AM

Thank you!
imo i think that solution is more easy than the first one -_-

Thanks.

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