anonymous
  • anonymous
Find the critical numbers of the function. h(t) = t^3/4 − 3t^1/4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
What did you get for the derivative?
anonymous
  • anonymous
I no im supposed to find the derivative
anonymous
  • anonymous
and solve for 0, but i didnt no how to solve for 0 for this equation

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anonymous
  • anonymous
i got 3/4t^-1/4-3/4t^-3/4
anonymous
  • anonymous
Looks right. Now set that = 0.
anonymous
  • anonymous
i did that but because of the exponenti get confused
anonymous
  • anonymous
\[\frac{3}{4t^{1/4}} - \frac{3}{4t^{3/4}} = 0\]
anonymous
  • anonymous
so i factor out 3/4t^-1/4?
anonymous
  • anonymous
I wouln't bother. Just set \(t^{1/4} = t^{3/4}\)
anonymous
  • anonymous
Cause that's the only way those two things can be equal.
anonymous
  • anonymous
\[\frac{3}{4}t^{-1/4} - \frac{3}{4}t^{-3/4} = 0\] \[\frac{3}{4}(t^{-1/4} - t^{-3/4}) = 0\] \[t^{-1/4} - t^{-3/4} = 0\] \[t^{-1/4} = t^{-3/4}\] \[t^{1} = t^{3}\] \[\implies t=1\]
anonymous
  • anonymous
how did u get t^1=t^3
anonymous
  • anonymous
raised both sides to the -4 power.
anonymous
  • anonymous
ooooo so it get crossed out
anonymous
  • anonymous
omg this way is much simpler then the way my teacher explained it!
anonymous
  • anonymous
thank u soo much!
anonymous
  • 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.
anonymous
  • anonymous
o ya i c that now

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