i just don't get this now...i got the x-coordinate, but maybe im doing something wrong to find the y coordinate :(

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

and how would i find the extremas of:

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

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- 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 :(

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

and how would i find the extremas of:

$\displaystyle y=x^3 + x^2 - 8x + 5$ - May 10th 2009, 04:05 PMpickslides
$\displaystyle y=x^3 + x^2 - 8x + 5$

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

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

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

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

Now solve for x using the quadratic formula.

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

sub these values back into y to get your answer...