1. ## 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.

2. 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 $\displaystyle (0,p)$
$\displaystyle \mathbb{Q}.\mathbb{E}.\mathbb{D}$