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anonymous
 3 years ago
HELP so frustrating!! Determine the extrema of below on the given interval
f(x)=5x^361x^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
 3 years ago
HELP so frustrating!! Determine the extrema of below on the given interval f(x)=5x^361x^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.

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can someone please tell me the correct answer to this. it seems like one of my answers is incorrect.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0check \(f(0), f(4)\) and also find the critical point and find \(f\) of that number

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a) 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
 3 years ago
Best ResponseYou've already chosen the best response.0the derivative is \( 15 x^2122 x+16\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which, by some miracle factors as \((x8) (15 x2)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0zeros of the derivative are \(\frac{2}{15}\) and \(8\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so \(\frac{2}{15}\) is the \(x\) coordinate of the local max and \(8\) is the \(x\) coordinate of the local min

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so for the intervals [0,4] max is: 2/15 and 8 is the min? what about [9,9]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i just want to make sure because it submitted the answer and it keeps saying its wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0any one! this is important!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0never mind! i found it. thanks to those who actually helped.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f'(x)=10x^2  122x +16 0=10x^2  122x + 16 (just working here because I don't have paper handy)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0careful here the max is the \(y\) value not the \(x\) value

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just finding critical points

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, sorry you already found them

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.08 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for \([9,9]\) you need to check \[f(9),f(\frac{2}{15}), f(8), f(9)\]
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