# Thread: Critical numbers, max & mins?

1. ## Critical numbers, max & mins?

So, I'm studying for an upcoming Calculus test, and I am stuck on only three problems.

1. Find the absolute or local maximum and minimum of f(x)=8-2x if x is greater than or equal to 6
• I pretty much just plugged 6 in for x and then 0 and.. yeah. I'm not really sure. I got -4 is an absolute minimum, but I doubt that's correct.

2. Find the critical numbers of f(x)=x^4(x-3)^3
• For this one, I found the derivative and was going to set it to 0, but I get a little confused. I end up with [4x^3(x-3)^3]+[3x^4(x-3)^2], and I don't know how to set that equal to 0 or if it can be simplified further.

3. Find the maximum or minimum of F(x)=(1-x^2)+6x^2
• Again, I found the first derivative and still got confused finding it (to set it to 0). I understand derivatives; it's just setting them equal to 0 that confuses me.

Any help will be greatly appreciated.

2. 1 is a downward sloping line so the max has to be 6 because the line decrease as you approach inifinity

$\displaystyle f'(x)=4x^3(x-3)^3+x^4*3(x-3)^2=x^3(x-3)^2*(4*(x-3)+x))=0$
Finishing simplifying.

$\displaystyle f'(x)=-2x+12x=0\Rightarrow 10x=0 \Rightarrow x=0$

Test numbers to the left and right of 0. When x=-1, the derivative is negative, and when x=1, it is positive. Therefore, at x=0, we have a min.

3. ## .

Thank you

So, basically,

1. is the absolute maximum at 6
2. I'm not sure
3. the point (0,1)?

4. 1. Where else could you have a maximum on a line that is strictly decreasing?

2. $\displaystyle x^3(x-3)^2*(4*(x-3)+x))=x^3(x-3)^2(4x-12+x)=x^3(x-3)^2(5x-12)=0\Rightarrow x^3=0, \ (x-3)^2=0, \ 5x-12=0$

5. ## .

Heh, I didn't mean to word number 1 as a question, I get it.

And, so for number 2, is the answer supposed to be 0, 3, and 12/7?

And the third one is the point (0,1)?

Thanks