Finding Relative max...

• December 8th 2009, 04:00 PM
penguinpwn
Finding 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

• December 8th 2009, 04:14 PM
Scott H
Because $f$ is differentiable everywhere and the domain $(0,3)$ contains no boundary points, relative maxima will only occur where

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

By a theorem in algebra, this occurs only where

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

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

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