Question

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.

Answer 1

can someone please tell me the correct answer to this. it seems like one of my answers is incorrect.

Answer 2

check \(f(0), f(4)\) and also find the critical point and find \(f\) of that number

Answer 3

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

Answer 4

the derivative is \( 15 x^2-122 x+16\)

Answer 5

which, by some miracle factors as \((x-8) (15 x-2)\)

Answer 6

zeros of the derivative are \(\frac{2}{15}\) and \(8\)

Answer 7

so \(\frac{2}{15}\) is the \(x\) coordinate of the local max and \(8\) is the \(x\) coordinate of the local min

Answer 8

so for the intervals [0,4] max is: 2/15 and 8 is the min?
what about [-9,9]

Answer 9

i just want to make sure because it submitted the answer and it keeps saying its wrong.

Answer 10

@satellite73

Answer 11

any one! this is important!

Answer 12

never mind! i found it. thanks to those who actually helped.

Answer 13

f'(x)=10x^2 - 122x +16
0=10x^2 - 122x + 16
(just working here because I don't have paper handy)

Answer 14

careful here
the max is the \(y\) value not the \(x\) value

Answer 15

just finding critical points

Answer 16

oh, sorry you already found them

Answer 17

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

Answer 18

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

Answer 19

got it. thank you!

Answer 20

yw