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 October 14th, 2013, 08:39 AM #1 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Find the critical numbers of the function. Find the critical numbers of the function. $h(t)= t^{3/4} - 2t{1/4}$ So I took the derivative. $h'(t) = \frac{3}{4}t^{-1/4} + \frac{1}{2}t^{-3/4} h'(t) = \frac{7}{4}t^{-1}$ ? Now what do I do? Make it a square root? Or? I'm confused. I need to set t = to zero to solve for the critical points correct?
 October 14th, 2013, 09:35 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Find the critical numbers of the function. For your first step, the sign of the second term is wrong, and I have no idea what you did for the second step. We are given the function: $h(t)=t^{\frac{3}{4}}-2t^{\frac{1}{4}}$ Differentiating with respect to $t$, we find: $h'(t)=\frac{3}{4}t^{-\frac{1}{4}}-\frac{1}{2}t^{-\frac{3}{4}}=\frac{3t^{\frac{1}{2}}-2}{4t^{\frac{3}{4}}}$ Now, can you proceed?
 October 14th, 2013, 10:40 AM #3 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: Find the critical numbers of the function. What do you mean by "critical points"? If you mean the local and absolute maxima and minima, then set the derivative to zero and solve for t.
 October 14th, 2013, 11:31 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Find the critical numbers of the function. No, the "critical numbers" are not the x values for max and min. The "critical numbers" of a function are values of x were the derivative is 0 or does not exist. That will include max and min points but, for example, x= 0 is a critical number for $f(x)= x^3$ but is neither a max nor a min. I thought every Calculus text gave that definition.

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### critical numbers of function of 4th degree

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