What does it mean to find the absolute extreme values? y= -5coshx + 4sinhx
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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?
do you make it equal to 0 and solve for x?
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
so itd be... y'= 5sinhx + 4coshx =0 ?
Originally Posted by runner07 so itd be... y'= 5sinhx + 4coshx =0 ? no, $\displaystyle \frac {d}{dx} \cosh x = \sinh x$
sorry, but i dont understand how you got that..
Originally Posted by Jhevon no, $\displaystyle \frac {d}{dx} \cosh x = \sinh x$ Originally Posted by runner07 sorry, but i dont understand how you got that.. $\displaystyle cosh(x) = \frac{e^x + e^{-x}}{2}$ So $\displaystyle \frac {d}{dx} \cosh x = \frac{d}{dx} \frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x}}{2} = sinh(x)$ -Dan
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