Question

Who's good at Calculus?

Answer 1

Consider the function f(x) = x^4 - 32 x^2 + 3,
-3 (less than or equal to) x (less than or equal to) 9. This function has an absolute minimum value equal to and an absolute maximum value equal to

Answer 2

1- find f'(x)
2- find x such that f'(x)=0
3- test what x represent ( max, min )

Answer 3

i got -4 and 4

Answer 4

\[-3\leq xleq 9\]does not contain the number \(-4\) so ignore that one

Answer 5

oops
\[-3\leq x\leq 9\]

Answer 6

so 4?

Answer 7

assuming your work is right, which i did not check, the only critical point in that interval would be 4 yes
check the critical point and the endpoint to find the max and min

Answer 8

how do you check it

Answer 9

second derivative

Answer 10

there are 3 critical points (one critical point is not in the interval though)

Answer 11

he forgot 0

Answer 12

second derivative is 12x^2-64

Answer 13

set that equal to zero and solve for x

Answer 14

no no

Answer 15

no?

Answer 16

also yadiel im on your cost question take a look

Answer 17

when looking at the second derivative you plug in your 0 and 4

Answer 18

since those are your critical points

Answer 19

if the answer is less than 0 then it is a maximum point
similarly if the answer is more than 0 it is a minimum point

Answer 20

This is entirely unnecessary. Just take the minimum over the critical points and the end points and the maximum over the critical points and end points.

Answer 21

We are looking for the absolute min/max on the interval.

Answer 22

You are looking for abs min and max, why 2nd derivative?

Answer 23

oh my hello @Alchemista !!

Answer 24

\(\min\ \{f(-3), f(0), f(4), f(9)\}\)
\(\max\ \{f(-3), f(0), f(4), f(9)\}\)

You do not need the second derivative test.

Answer 25

ah.. you are right

Answer 26

just plugging in points to original equation is good

Answer 27

Don't forgot the endpoint! alright