Question

Find the critical numbers of the function.
h(t) = t^3/4 − 3t^1/4

Answer 1

What did you get for the derivative?

Answer 2

I no im supposed to find the derivative

Answer 3

and solve for 0, but i didnt no how to solve for 0 for this equation

Answer 4

i got 3/4t^-1/4-3/4t^-3/4

Answer 5

Looks right. Now set that = 0.

Answer 6

i did that but because of the exponenti get confused

Answer 7

\[\frac{3}{4t^{1/4}} - \frac{3}{4t^{3/4}} = 0\]

Answer 8

so i factor out 3/4t^-1/4?

Answer 9

I wouln't bother. Just set \(t^{1/4} = t^{3/4}\)

Answer 10

Cause that's the only way those two things can be equal.

Answer 11

\[\frac{3}{4}t^{-1/4} - \frac{3}{4}t^{-3/4} = 0\]
\[\frac{3}{4}(t^{-1/4} - t^{-3/4}) = 0\]
\[t^{-1/4} - t^{-3/4} = 0\]
\[t^{-1/4} = t^{-3/4}\]
\[t^{1} = t^{3}\]
\[\implies t=1\]

Answer 12

how did u get t^1=t^3

Answer 13

raised both sides to the -4 power.

Answer 14

ooooo so it get crossed out

Answer 15

omg this way is much simpler then the way my teacher explained it!

Answer 16

thank u soo much!

Answer 17

But really you coulda guessed t would be 1 looking at the original equation for the derivative because

\[\frac{3}{4t^a} - \frac{3}{4t^{3a}} = 0\]
Will only be true if \(t^a = t^{3a}\) which means that t must be 1.

Answer 18

o ya i c that now