## anonymous 5 years ago Find the critical numbers of the function. h(t) = t^3/4 − 3t^1/4

1. anonymous

What did you get for the derivative?

2. anonymous

I no im supposed to find the derivative

3. anonymous

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

4. anonymous

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

5. anonymous

Looks right. Now set that = 0.

6. anonymous

i did that but because of the exponenti get confused

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

how did u get t^1=t^3

13. anonymous

raised both sides to the -4 power.

14. anonymous

ooooo so it get crossed out

15. anonymous

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

16. anonymous

thank u soo much!

17. anonymous

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.

18. anonymous

o ya i c that now