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.

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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.

Calculus1
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Do you know the derivative?
yes i did everything. x= 2/15 and x= 8
those are the critical points.

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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
Okay you don't consider \(f(8)\) for the \([0,4]\) interval..
oh......
It's apparent that \(f(2/15)\) is the maximum.
And the minimum is going to be \(f(4)\)
just 2/15? or do we write that the max as: 4.06074
so the min has to be: -589
I said \(f(2/15)\) which is equal to \(4.06074\)
okay. then what's the minimum
Now you just need to check 9 and -9 for the other interval.
since i can't use f(8)
For the \([0,4]\) interval, the minimum is -589 right?
i think so...
For the \([-9,9]\) interval... you can consider \(f(8)\), but you also need to consider \(f(-9)\) and \(f(9)\)
after i do all the work, the problem is i don't know how to pick the min and max.
for f(-9)= -8727 and for f(9) = -1149.
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
Yeah... the minimum is the LOWEST value the maximum is the HIGHEST value.
You only have to consider the values at critical points and endpoints.
i submitted this as my answer but it says its wrong. :(
You must be doing it wrong then... somehow.
i know that's why im asking you if the answer i had up there are correct.
What's your derivative?
15x^2-122x+16
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...

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