# Calc Question

• Mar 19th 2006, 04:24 PM
Vigo
Calc Question
Let f be the function given by f(x)= (x^3)-(6x^2)+p, where p is an arbitrayry constant.

a)Write an expression for f'(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.

b)For what values of the constant p does f have 3 distinct roots?

c)Find the value of p such that the average value of f over the closed interval [-1,2] is 1.

For a), I found the derivative, which is f'(x)=(3x^2)-12x. I set that equal to 0, which gives me x=0,4. I put these values back into the original equation and got f(0)=o and f(4)=-32. In terms of p, I got 0=p and 32=p. Is this right?

I have no clue how to do b). I kind of know how to do c) but I would appreciate some help with it. Thanks a lot.
• Mar 19th 2006, 07:12 PM
ThePerfectHacker
Quote:

Originally Posted by Vigo
Let f be the function given by f(x)= (x^3)-(6x^2)+p, where p is an arbitrayry constant.

a)Write an expression for f'(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.

\$\displaystyle f(x)=x^3-6x^2+p\$
\$\displaystyle f'(x)=3x^2-12x\$
\$\displaystyle f''(x)=6x-12\$
Find the critical points, \$\displaystyle f'(x)=0\$ thus,
\$\displaystyle 3x^2-12x=0\$
\$\displaystyle x(3x-12)=0\$
\$\displaystyle x=4,0\$
But, \$\displaystyle f''(0)<0 \mbox{ and }f''(4)>0\$
Thus, \$\displaystyle x=0\$ is maximum (2nd derivative test)
and \$\displaystyle x=4\$ is minimum.
But, \$\displaystyle f(0)=p\$ and \$\displaystyle f(4)=-32+p\$
Thus, min at \$\displaystyle (4,-32+p)\$ max at \$\displaystyle (0,p)\$
\$\displaystyle \mathbb{Q}.\mathbb{E}.\mathbb{D}\$