How do I find the relative max on the interval 0<(or equal) X <(or equal) 3

when given the first derivative as:

f ' (x) = (e^(-x/4))(sinx^2)

where x is greater than or equal to zero

& f(5)=0

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- Dec 8th 2009, 03:00 PMpenguinpwnFinding Relative max...
How do I find the relative max on the interval 0<(or equal) X <(or equal) 3

when given the first derivative as:

f ' (x) = (e^(-x/4))(sinx^2)

where x is greater than or equal to zero

& f(5)=0

- Dec 8th 2009, 03:14 PMScott H
Because $\displaystyle f$ is differentiable everywhere and the domain $\displaystyle (0,3)$ contains no boundary points, relative maxima will only occur where

$\displaystyle f'(x)=e^{-\frac{x}{4}}\sin x^2=0.$

By a theorem in algebra, this occurs only where

$\displaystyle e^{-\frac{x}{4}}=0\quad\mbox{or}\quad \sin x^2=0.$

Because $\displaystyle e^{-\frac{x}{4}}$ is always positive, we are left with the cases in which

$\displaystyle \sin x^2=0\quad\quad\quad 0 < x < 3.$