# more extreme values

• May 10th 2009, 02:56 PM
>_<SHY_GUY>_<
more extreme values
i just don't get this now...i got the x-coordinate, but maybe im doing something wrong to find the y coordinate :(

$y=2x^x-8x+9$

and how would i find the extremas of:
$y=x^3 + x^2 - 8x + 5$
• May 10th 2009, 04:05 PM
pickslides
Quote:

Originally Posted by >_<SHY_GUY>_<
$y=2x^x-8x+9$

and how would i find the extremas of:
$y=x^3 + x^2 - 8x + 5$

$y=x^3 + x^2 - 8x + 5$

find extremas of this function make the derivative equal to zero.

$\frac{dy}{dx}=3x^2 + 2x - 8$

$\frac{dy}{dx}=0$

$3x^2 + 2x - 8 = 0$

Now solve for x using the quadratic formula.

$x = \frac{-2\pm\sqrt{2^2-4\times3\times-8}}{2\times3}$