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H-E-L-P so frustrating!! 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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can someone please tell me the correct answer to this. it seems like one of my answers is incorrect.
check \(f(0), f(4)\) and also find the critical point and find \(f\) of that number
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

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Other answers:

the derivative is \( 15 x^2-122 x+16\)
which, by some miracle factors as \((x-8) (15 x-2)\)
zeros of the derivative are \(\frac{2}{15}\) and \(8\)
so \(\frac{2}{15}\) is the \(x\) coordinate of the local max and \(8\) is the \(x\) coordinate of the local min
so for the intervals [0,4] max is: 2/15 and 8 is the min? what about [-9,9]
i just want to make sure because it submitted the answer and it keeps saying its wrong.
any one! this is important!
never mind! i found it. thanks to those who actually helped.
f'(x)=10x^2 - 122x +16 0=10x^2 - 122x + 16 (just working here because I don't have paper handy)
careful here the max is the \(y\) value not the \(x\) value
just finding critical points
oh, sorry you already found them
8 is not in the interval \([0,4]\) for that one, you need to check \(f(0), f(4), f(\frac{2}{15})\) the largest is the max and the smallest is the min
for \([-9,9]\) you need to check \[f(-9),f(\frac{2}{15}), f(8), f(9)\]
got it. thank you!

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