Question

Find the absolute maximum value for the function f(x) = x^2 - 3, on the interval [-4, 0) U (0, 3]. (2 points)

3
4
13
No maximum exists because f is not continuous


Idk how to do this, plz include steps on how to get the answer, thank you! :)

Answer 1

"the absolute maximum value for a function" could be calculated from its first derivative by equating the first derivative to zero and find the zeros of this derived function!

Answer 2

So the first derivative would be -32t and if it equals 0, t would be equal 0 so the answer would be D?

Answer 3

"would be -32t"
what \(t\) ?
\[\Huge f(x) = x^2 - 3\]
that is our equation................................

Answer 4

>.> i'm confused o-o

Answer 5

Why?

The given function is \[\Huge f(x) = x^2 - 3\]
and its first derivative is
\[\Huge f'(x) = \frac{ d }{ dx }(x^2 - 3)=2x\]

Answer 6

sorry i'm confusing questions >.<
but how do you find a max on a line?

Answer 7

oh wait... nvm heh it doesn't exist cuz it's a line

Answer 8

Let me simplify it to you..

you calculated \(f'(x)\) and it is \(2x\),,,what did you get after equating it to zero?

Answer 9

that would be 2x=0 so x=0

Answer 10

You would get that
\[2x=0\\x=0\]

this value of x=0 means that the slope of the function is undefined or zero..
"Critical points of f are when f'(x)=0 or when f'(x) is undefined"

i.e x=0 is the critical point of f(x).

Answer 11

So the answer is D?

Answer 12

Step by step....

After getting that critical point \(x=0\), search if it is in the specified domain or not.

Is in is or out of it?

Answer 13

@volpina

\(x=0\) lies in the given domain or it is not included?

Answer 14

It's included because both ranges include 0

Answer 15

[-4, 0) U (0, 3] does that mean \(x=0\) included or not?

Answer 16

Wait, it's not because zero doesn't have a square bracket

Answer 17

Yes, That is what I want you to take care about.