# Critical Number Problem

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• October 17th 2013, 12:03 PM
Jason76
Critical Number Problem
Find the critical numbers:

$h(t) = t^{3/4} - 3t^{1/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$ = 0 http://www.freemathhelp.com/forum/im...s/confused.png Next move. How can I solve for t in this situation?
• October 17th 2013, 12:14 PM
SlipEternal
Re: Critical Number Problem
Multiply both sides by $t^a$ where $-a$ is the smallest exponent of $t$. (In other words, multiply both sides by $t^{3/4}$).
• October 17th 2013, 12:24 PM
Jason76
Re: Critical Number Problem
Post Edited

Find the critical numbers:

$h(t) = t^{3/4} - 3t^{1/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$ = 0

$h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}$

$h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}$ http://www.freemathhelp.com/forum/im...n_question.gif
• October 17th 2013, 12:44 PM
SlipEternal
Re: Critical Number Problem
Quote:

Originally Posted by Jason76
Post Edited

Find the critical numbers:

$h(t) = t^{3/4} - 3t^{1/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$ = 0

$h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}$

$h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}$ http://www.freemathhelp.com/forum/im...n_question.gif

Huh? Why would you think $t^{-3/4} = t^{-4/4}$?
• October 17th 2013, 06:20 PM
Jason76
Re: Critical Number Problem
Find the critical numbers:

$h(t) = t^{3/4} - 3t^{1/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$ = 0

$h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}$

$h'(t) = t^{-3/4} = t^{-1/4}$

$h'(t) = \dfrac{ t^{-1/4}}{t^{-3/4}} = 0$

$h'(t) = t^{1/2} = 0$ ??
• October 17th 2013, 06:33 PM
SlipEternal
Re: Critical Number Problem
Why do you keep writing $h'(t)$? When you go from the line that reads
$h'(t) = \dfrac{3}{4}t^{-1/4}-\dfrac{3}{4}t^{-3/4} = 0$
to the line that reads
$h'(t) = -\dfrac{3}{4}t^{-3/4} = -\dfrac{3}{4}t^{-1/4}$, you no longer have $h'(t)$. You subtracted $\dfrac{3}{4}t^{-1/4}$ from every term, so you should have $h'(t) - \dfrac{3}{4}t^{-1/4} = -\dfrac{3}{4}t^{-3/4} = -\dfrac{3}{4}t^{-1/4}$

So just drop $h'(t) =$ from every line. It is not true after line 5.

Next, why would you think that $\dfrac{t^{-3/4}}{t^{-3/4}} = 0$?
• October 17th 2013, 06:42 PM
Jason76
Re: Critical Number Problem
Find the critical numbers:

$h(t) = t^{3/4} - 3t^{1/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$

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

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

$t^{-3/4} = t^{-1/4}$

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

$t^{1/2} = 0$
• October 17th 2013, 06:46 PM
SlipEternal
Re: Critical Number Problem
Also, from now on, it might be easier to help you if you number each line. I am "quoting" you, but I will make changes to your quote to add line numbers and make each line correct.

Quote:

Originally Posted by Jason76
Find the critical numbers:

1. $h(t) = t^{3/4} - 3t^{1/4}$

2. $h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}$

3. $h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}$

4. $h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$

5. $h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}=0$

6. $-\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}$

7. $t^{-3/4} = t^{-1/4}$

8. $\dfrac{ t^{-1/4}}{t^{-3/4}} = 0$ <-- This line is wrong

9. $t^{1/2} = ???$

Now, the question is, how did you get from line 7 to line 8?
• October 17th 2013, 10:02 PM
votan
Re: Critical Number Problem
Quote:

Originally Posted by Jason76
Post Edited

Find the critical numbers:

$h(t) = t^{3/4} - 3t^{1/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}$

$h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$

$h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}$ = 0

$h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}$

$h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}$ http://www.freemathhelp.com/forum/im...n_question.gif

You are giving multiple = sign in the same line without showing how you got the right side from the left side. Help us understand what you are doing by writing it. We cannot decipher what you have in mind what you write your expression the way you do. Perhaps you need to write one = sing in a line, and write a short comment next to the line telling what did you do to get from one line to the next.