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So $h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$. So I need to set $h'(t)=0$. So for $h'(t)$ the fattest I've gotten to simplifying os $h'(t)=\frac{3}{4 \sqrt[4]{t}}-\frac{7}{4\sqrt[4]{t^3}}$ and that is as farthest as I can simplify. So i'm having a had time having $h'=0$ So could someone show me how to properly simplify and find the critical values step by step I would immensely appreciate it. In the most recent material we have covered in class ir seems that my biggest struggle comes from simplifying completely. Thanks in advance for the help.

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Pull out a factor of $\frac{1}{4\sqrt[4]{t^3}}$. You know that isn't $0$, so the other factor must be $0$. The other factor looks simpler. –  Daniel Fischer Jul 12 at 23:36

3 Answers 3

up vote 7 down vote accepted

$$h'(t)=\frac{3}{4 t^{\frac{1}{4}}}-\frac{7}{4t^{\frac{3}{4}}}=\frac{3t^{\frac{2}{4}}-7}{4t^{\frac{3}{4}}}=\frac{3t^{\frac{1}{2}}-7}{4t^{\frac{3}{4}}}$$

$$h'(t)=0 \Rightarrow 3t^{\frac{1}{2}}-7=0 \Rightarrow t^{\frac{1}{2}}=\frac{7}{3} \Rightarrow t=\frac{49}{9}$$

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Thanks a lot, but I am having trouble understanding how you went from $\frac{3}{4 t^{\frac{1}{4}}}-\frac{7}{4t^{\frac{3}{4}}}->\frac{3t^{\frac{2}{4}}-7} {4t^{\frac{3}{4}}}$ If you could explain I'd really appreciate it. Thanks in advance. –  Kenshin Jul 12 at 23:58
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@Kenshin Multiply the first fraction by $\frac{t^\frac{1}{2}}{t^{\frac{1}{2}}}$. Now in that denominator $4t^{\frac{1}{4}}t^\frac{1}{2}=4t^{\frac{3}{4}}$, and you can join the two fractions as one. –  J. W. Perry Jul 13 at 0:01
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Thanks a lot @J.W.Perry! Appreciate the help. –  Kenshin Jul 13 at 0:20

$h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$

$\Rightarrow$ $0=h'(t)=\frac{3}{4}t^{-\frac{1}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}+\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}}t^{\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=t^{-\frac{3}{4}}(\frac{3}{4}t^{\frac{2}{4}}-\frac{7}{4})$

Thus $t=(\frac{7}{3})^2$ Since one factor have to be $0$ and $t^{-\frac{3}{4}}$ is impossible.

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Hint

To make your life easier, just define $x=t^{\frac{1}{4}}$. So $h(t)=x^3-7x$ and then $$\frac{dh}{dt}=\frac{dh}{dx}\frac{dx}{dt}=(3x^2-7)\frac{dx}{dt}$$ So,$h'(t)=0$ if $x^2=\frac{7}{3}$ that is to say $\sqrt t=\frac{7}{3}$ then $t=\frac{49}{9}$.

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