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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

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 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

for \([-9,9]\) you need to check
\[f(-9),f(\frac{2}{15}), f(8), f(9)\]

got it. thank you!

yw