Question

Determine the critical values: 2x^2-64sqrt(x)

Answer 1

Critical values occur when the slope is zero. Batman can take it away if he so desires! =P

Answer 2

ok i know that but the square root is throwing me

Answer 3

Show me your best attempt. Also, you can rewrite a square root as:\[\sqrt{x}=x^{1/2}\] Which makes sense with the rest of exponent rules. =)

Answer 4

i got that but then what?

Answer 5

Show me what you got.

Answer 6

Do you know the definition of a critical value?

Answer 7

yes the derivative = 0
I got the derivative: 4x-32sqrtx/x

Answer 8

i just dont know how to slove for 0

Answer 9

ok, here is the definition of a critical value.
c is call a critical value of a function f if f'(c) = 0 or f'(c) does not exist.

So there are two things you need to find: where f has slope = 0, or where f' does not exist.

First thing is you need to take the derivative, which you got it incorrect
f'(x) = 4x - 32x^(-1/2)

Answer 10

oh yeah i just rationalized the denominator

Answer 11

:o oh, didn't realize that. Ok, so x = 0 is certain one of the critical value since f'(0) does not exist.

the other one is f'(x) = 0
4x-32sqrt(x)/x = 0

then solve for x.

Answer 12

https://www.wolframalpha.com/input/?i=4x-32sqrt%28x%29%2Fx+%3D+0
so x = 0 or 4 <=== ans

Answer 13

so how did ya get x=0?

Answer 14

does f'(0) exist?

Answer 15

in other words, can you do
f'(0) = 4x - 32sqrt(0)/0 ?

Answer 16

C is also in the domain of f(x) not sure if you mentioned that, but just pointing it out.

Answer 17

but how does the derivative=0 come out to be 0? I got the 4, but not the 0

Answer 18

Set your derivative = 0

Answer 19

You have to check where f'(x) = 0 and where f'(x) = undefined.

Answer 20

x = 4 makes f' = 0.

But the fact that x = 0 is also a critical value is because f'(0) does not exist. and 0 is also in the domain of f as iambatman mentioned

Answer 21

So if you can show your work, it'll be easier to guide you.

Answer 22

ok so4x-32x^.5=0
4x-32/sqrtx=0
4x=32/sqrtx
32=4xsqrtx
8=x^3/2
x=4
but i dont see how it can be 0 from this anyways :)

Answer 23

again, if you look at the definition, 0 didn't come out from solving for x of the equation f'(x) = 0. It comes from the fact that 0 is in the domain of f, and f'(0) does not exist.

Answer 24

ok so did i solve for 0 right?

Answer 25

you mean if you found x = 0 the right way?

Answer 26

ya

Answer 27

you only found x = 4.

Answer 28

so i guess my question is,, is there a way to find the x=o like i found the x=4? sorry i confused you

Answer 29

You get 0 from seeing where f'(x) is undefined.

Answer 30

yes. Most of the times, it comes from the denominator because the's where division by 0 occurs.

recall that f'(x) = 4x - 32sqrt(x)/x. Whatever the denominator is, you set it equals to 0 and solve.

since x is is the only term that is in the denominator, setting it equal to 0, i.e x = 0, which gives you x = 0 when you solve

Answer 31

oh ok i see it halleluja! haha thanks!!! both of you!

Answer 32

np

Answer 33

These are the main things you should remember when looking for critical points.

f'(x) = 0
f'(x) = undefined
c is in the domain of f(x)

If I'm missing something please let me know @sourwing XD

Answer 34

@iambatman that's correct. those are the only 3 criteria to find the critical values.

Answer 35

Thought so, just checking :p