This is a practice exam problem am stuck on
Find the max and min values of the fuction f(x)=x^3-9x^2+15x+3 on intervals [0,2]
The global extrema of a function on a closed interval are either calculus type local extrema in the interior of the interval or occur at the end points of the interval.
So first look at the solutions of:
$\displaystyle f'(x)=3x^2-18x+15=0$
these are the roots of $\displaystyle x^2-6x+5=0$, which are $\displaystyle x=5$ and $\displaystyle x=1$, $\displaystyle x=5$ is not in the interval so we can ignore it, so now we need only consider
$\displaystyle f(0),\ f(1)$ and $\displaystyle f(2),$
$\displaystyle f(0)=3,$
$\displaystyle f(1)=1-9+15+3=10$
$\displaystyle f(2)=8-36+30+3=5$
so the global minimum on $\displaystyle [0,2]$ is $\displaystyle 3$ and the global maximum is $\displaystyle 10$ and these occur when $\displaystyle x=0$ and $\displaystyle x=1$ respectivly.
RonL