Question

Cal I question, I'm stuck: So I'm trying to find absolute values of: f(x) =3x^4-4x^3-12x^2+1. I found its' derivative, and found the critical values to be 0,2,-1. My interval is [-2,3]. I plugged those five values in to my original function, and only one landed within the interval...where does that leave me?

Answer 1

0, 2 and -1 are not the only critical values

Answer 2

is f(x) = abs(3x^4-4x^3-12x^2+1) ??

Answer 3

No, it's not within an absolute value?

Answer 4

what is within the absolute value?

Answer 5

Where did this absolute value come from? I didn't write it in the question o.o

Answer 6

you said you're trying to find the absolute value of f(x)

Answer 7

Ugh, my God xD. I meant ABSOLUTE MAX AND MINS. SORRY XD

Answer 8

Absolute maxima and minima of the function, my bad xP

Answer 9

don't you mean local max and local min? f(x) opens up, so there is no absolute max. But there will be an absolute min

Answer 10

why not?

Answer 11

No, the question states Find the absolute maximum and minimum values of f(x) on [-2,3]

Answer 12

abs min f(2) = -47
abs max f(-2) =33

Answer 13

:o so [-2,3] is the domain. I thought you came up with it somewhere :DD

Answer 14

No, I didn't. It's the interval they gave me. Sorry for not being more clear xP

Answer 15

@Loser66 But doesn't it have to be within the interval??

Answer 16

the interval, itself is critical point

Answer 17

Then discard the ones that are not in the interval

Answer 18

Are you supposed to plug it into the original equation or the derivative? Because f(2) and f(-2) gave me -7 and 33 respectively. And that was my original question @sourwing That if I only have ONE value in the interval, what does that make it?

Answer 19

but 0,-1,2 are in [-2,3].

So you need to check f(0), f(-1), f(2), f(-2) and f(3). Whichever is the smallest is the absolute min, and whichever is largest is the absolute max

Answer 20

Ah, okay, I must have confused myself. I thought that whatever f(0), f(-1), etc...gave must be within in the interval.

Answer 21

those are the values of the function within [-2,3]

Answer 22

remember [-2,3] is the values of x, while f(0), f(-), ... are values of f

Answer 23

Okay, so just to be clear, it's the critical points I calculate that must be within in the interval, and the highest/lowest values of those (once plugged in) would be the abs max/min?

Answer 24

yes, critical values must be within the given interval. but f(3) isn't necessarily bigger than f(-2) just because 3 > -2. You just have to check.

Answer 25

I understand that. Just wanted to be sure the values once plugged in don't have to be in the interval. Thank you to the both of you.

Answer 26

np