H-E-L-P Determine the extrema of below on the given interval
f(x)=5x^3-61x^2+16x+3

(a) on [0,4]
The minimum is ?? and the maximum is ??
(b) on [-9,9]
The minimum is ?? and the maximum is ??

please solve so i can know which answer i got wrong.

Question

H-E-L-P Determine the extrema of  below on the given interval
 f(x)=5x^3-61x^2+16x+3

(a) on [0,4]
The minimum is ?? and the maximum is ??
(b) on [-9,9] 
The minimum is ??  and the maximum is ??

please solve so i can know which answer i got wrong.

anonymous · Answer

Do you know the derivative?

anonymous · Answer

yes i did everything. x= 2/15 and x= 8

anonymous · Answer

those are the critical points.

anonymous · Answer

left end point: 
f(0)= 3 

critical points:
 x= 2/15 f(2/15)= 4.06074
 x=8 f(8)= -1213 

Right Point: 
f(4)= -589

anonymous · Answer

Okay you don't consider $f(8)$ for the $[0,4]$ interval..

anonymous · Answer

oh......

anonymous · Answer

It's apparent that $f(2/15)$ is the maximum.

anonymous · Answer

And the minimum is going to be $f(4)$

anonymous · Answer

just 2/15? or do we write that the max as: 4.06074

anonymous · Answer

so the min has to be: -589

anonymous · Answer

I said $f(2/15)$ which is equal to $4.06074$

anonymous · Answer

okay. then what's the minimum

anonymous · Answer

Now you just need to check 9 and -9 for the other interval.

anonymous · Answer

since i can't use f(8)

anonymous · Answer

For the $[0,4]$ interval, the minimum is -589 right?

anonymous · Answer

i think so...

anonymous · Answer

For the $[-9,9]$ interval... you can consider $f(8)$, but you also need to consider $f(-9)$ and $f(9)$

anonymous · Answer

after i do all the work, the problem is i don't know how to pick the min and max.

anonymous · Answer

for f(-9)= -8727 and for f(9) = -1149.

anonymous · Answer

Is this correct ??

(a) on [0,4]
The minimum is -589 and the maximum is 4.06074

(b) on [-9,9] 
The minimum is -8727 and the maximum is -1149

anonymous · Answer

Yeah... the minimum is the LOWEST value   the maximum is the HIGHEST value.

anonymous · Answer

You only have to consider the values at critical points and endpoints.

anonymous · Answer

i submitted this as my answer but it says its wrong. :(

anonymous · Answer

You must be doing it wrong then... somehow.

anonymous · Answer

i know that's why im asking you if the answer i had up there are correct.

anonymous · Answer

What's your derivative?

anonymous · Answer

15x^2-122x+16

anonymous · Answer

can you get back to me with answer so i can see if its correct. i dont want to go step by step if its wrong in the end. i tried this problem several times...