Question

Find the critical points of the function.

Answer 1

h(t) = t^(3/4) - 2t^(1/4)

Answer 2

I know how to do these, but I got the wrong answer for this one for some reason. First I took the derivative:
3/4(t)^(-1/4) - 1/2(t)^(-3/4)

Sorry, I don't know how to do fractional exponents on here :P
So then I tried to set it equal to zero... guess that's where I got stumped

Answer 3

use radicals

Answer 4

and also subtract, so you have one fraction to set the numerator equal to zero

Answer 5

Yep, that's what I did. I moved the negative exponents to the bottom of a fraction.. like dis:
3 over 4*fourthroot(t) - 1 over 2*fourthroot(t)^3

How would I subtract that bad boy? D:

Answer 6

common denominator is \[4\sqrt[4]{t^3}\]

Answer 7

Okay, so multiply the second fraction by 2/2... remind me how to do the first fraction...

Answer 8

Multiply by fourthroot(t^2) ?

Answer 9

you have to multiply \[\frac{3}{4\sqrt[4]{t}}\] by \(\frac{\sqrt[4]{t^2}}{\sqrt[4]{t^2}}\)

Answer 10

which is, needles to say, the same as \[\frac{\sqrt t}{\sqrt t}\]

Answer 11

let me know what you get in the numerator

Answer 12

The numerator will just be \[3\sqrt[4]{t^2}\] right?

Answer 13

after you subtract it will be \[3\sqrt{t}-2\] but yeah

Answer 14

you do know that \[\sqrt[4]{t^2}=\sqrt{t}\] right?

Answer 15

Ah, yes

Answer 16

Would I set the numerator to zero now?

Answer 17

so now \[3\sqrt{t}-2=0\] should be routine

Answer 18

yup

Answer 19

Gotcha, thanks so much for the help!

Answer 20

you also have a critical point a 0, the zero of the denominator, but that is no surprise since it is the endpoint of your domain

Answer 21

yw

Answer 22

Why would zero also be a critical point? I thought zeros in the denominator made the function undefined and didn't count as critical points?

Answer 23

critical points are where the derivative is zero, or where it does not exist

Answer 24

in practice that usually means you take the derivative and find the zeros of the denominator

Answer 25

Oh, so when I have fractional derivatives should I be setting both the numerator and denominator equal to zero to find my CVs?

Answer 26

Oh okay

Answer 27

exactly
the zero of the denominator of the derivative
notice that your function is defined at 0, but your derivative is not

Answer 28

likewise for example \[f(x)=\sqrt[3]{x}\] exists everywhere but the derivative is not defined at zero

Answer 29

\[h'(t)=\frac{ 3 }{ 4\sqrt[4]{t} }-\frac{ 1 }{ 2\sqrt[4]{t^{3}} }\]
Now solve h'(t)=0 to find the critical points.

Answer 30

Isn't that what we just did?