Analyzing f'(x) to Find Relative Max/Min of f in Terms of p

[Vigo]

Let f be the function given by f(x)= (x^3)-(6x^2)+p, where p is an arbitrary 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.

 

[Stephan Hoyer]

a) Your maximum and minimum values of f (f(0) and f(4)) should be in terms of p, as specified in the problem. p is a constant that is the same for any given case of the function. You have no requirement that f(x)=0, so you can choose whatever values of p you desire.

b) Try graphing f with p=0, and consider how changing the value of p would affect the graph.

 

[Vigo]

For a), I've got the points in terms of p.

For b), I'm just not quite sure what they are asking for when they say "distinct roots."

For c), I know that to find the average value, you take 1/(-1-2) and multiply that by the definite integral of f'(x) from -1 to 2. Since it has to equal 1, do 1=(-1/3)(?) I am not sure what to put where the question mark is.

 

[HallsofIvy]

For a), I've got the points in terms of p.

No, you don't. For some reason, you calculated f(0) and f(4), assuming p was 0, and then set that equal to p?
f(x)= x³- 6x²+ p. What is f(0)? What is f(4)? Do you see that the answer to those question still has "p" in it?

For b), I'm just not quite sure what they are asking for when they say "distinct roots."

"roots"- solutions to the equation. "distinct"- different.
A cubic equation such as x³- 6x²+ p= 0 may have 3 'distinct' solutions or it may have a "double root" or a triple root. What values of p guarantee that this equation does not have a double or triple root? (Hint: if a, b, c are distinct roots of polynomial Q(x) then Q(x)= d(x-a)(x-b)(x-c) so Q'(x)= d(x-b)(x-c)+ d(x-a)(x-c)+ d(x-a)(x-b). Setting x= a, b, c does NOT make that 0. If Q has a as double root and other root b then Q(x)= (x-a)[suQ]2[/suQ](x-b) so Q'(x)= 2(x-a)(x-b)+ (x-a)[suQ]2[/suQ] and Q'(a)= 0. If Q has a as triQle root, then Q(x)= (x-a)[suQ]3[/suQ] so Q'(x)= 3(x-a)[suQ]2[/suQ] and again Q'(a)= 0.) What values of Q guarantee that the same x value cannot make both Q and Q' equal to 0 (what values of x make Q'= 0? What is the value of Q(x) for those values of x?)

For c), I know that to find the average value, you take 1/(-1-2) and multiply that by the definite integral of f'(x) from -1 to 2. Since it has to equal 1, do 1=(-1/3)(?) I am not sure what to put where the question mark is.

Okay, so go ahead and do the integral! What is
$$\int_{-1}^2 (x^3- 6x^2+ p)dx$$
THAT'S what you want.