absolute extreme values?

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September 8th 2007, 09:01 PM
runner07

absolute extreme values?

What does it mean to find the absolute extreme values?

y= -5coshx + 4sinhx
September 8th 2007, 09:03 PM
Jhevon

Quote:

Originally Posted by runner07

What does it mean to find the absolute extreme values?

y= -5coshx + 4sinhx

find the extreme values. the largest one (the absolute max) and the smallest one (the absolute min) are the ones you're after.

you do remember how to find extreme values right?
September 8th 2007, 09:14 PM
runner07

do you make it equal to 0 and solve for x?
September 8th 2007, 09:17 PM
Jhevon

Quote:

Originally Posted by runner07

do you make it equal to 0 and solve for x?

you set the derivative = 0 and solve for x, and state the f(x) for that (or those) x's. you also have to check the end points as well
September 9th 2007, 10:48 AM
runner07

so itd be...

y'= 5sinhx + 4coshx =0 ?
September 9th 2007, 11:05 AM
Jhevon

Quote:

Originally Posted by runner07

so itd be...

y'= 5sinhx + 4coshx =0 ?

no, $\frac {d}{dx} \cosh x = \sinh x$
September 9th 2007, 11:07 AM
runner07

sorry, but i dont understand how you got that..
September 9th 2007, 11:23 AM
topsquark

Quote:

Originally Posted by Jhevon

no, $\frac {d}{dx} \cosh x = \sinh x$

Quote:

Originally Posted by runner07

sorry, but i dont understand how you got that..

$cosh(x) = \frac{e^x + e^{-x}}{2}$

So
$\frac {d}{dx} \cosh x = \frac{d}{dx} \frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x}}{2} = sinh(x)$

-Dan

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