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