# minimum values

• Jun 11th 2008, 02:41 AM
Math Phobe
minimum values
This is a practice exam problem am stuck on(Headbang)

Find the max and min values of the fuction f(x)=x^3-9x^2+15x+3 on intervals [0,2]
• Jun 11th 2008, 02:58 AM
CaptainBlack
Quote:

Originally Posted by Math Phobe
This is a practice exam problem am stuck on(Headbang)

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