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I'm stuck on this problem involving open intervals and absolute maximum/minimum.

$H(t)=t^6-\frac{3}{4}t^8$ On what intervals is the function increasing? Decreasing? What are the absolute maximum? Minimum?

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    Here is a [related problem](http://math.stackexchange.com/questions/173604/monotonocity-of-the-function/173622#173622).2012-11-12

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$H^{\prime}(t)=6t^5-8\cdot \frac{3}{4}t^7 = 6t^5 - 6t^7 = 6t^5(1-t^2) = 6t^5(1-t)(1+t).$

When is $H^{\prime}(t) = 0$?

  • These are points you need to look at to determine the maximum and minimum values of $H(t)$. There are three distinct values of $t$ at which $H^{\prime}(t) = 0$.

On what intervals is the function increasing? Decreasing?

  • Examine the behavior of $H(t)$ within each of the four intervals separated by the points at which $H^{\prime} = 0$.
  • On which intervals is $H^{\prime}(t)>0\;$? On these interval(s), $H(t)$ is increasing.
  • On which intervals is $H^{\prime}(t) <0\;$? On these intervals, $H(t)$ is decreasing.

Recall, the derivative of a function corresponds to the "rate of change" of the function.


To "see" the function's behavior, see $H(t)$ on WolframAlpha. Note the "action" going on in the interval, say $x\in (−1.5,1.5)$ that might easily get overlooked when graphed, depending on the scale of the $x$-$y$ axes.

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    Nice analysis and CONGRATULATIONS on 50K!!!! Well deserved! +12013-05-05
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The sign of the derivative $H^\prime(t)$ tells you where the function is increasing and decreasing. If $H^\prime>0$, it's increasing. If $H^\prime<0$, it's decreasing. So, the first thing to do is find $H^\prime$. Then, make a "sign chart" for $H^\prime$ - basically a number line that shows where $H^\prime(t)$ is positive, negative and zero. Below is an example of a sign chart (this isn't the sign chart for your function!!) This isn't the sign chart for your function, just an example of what it should look like!

You find these signs by 'testing' points between your zeros. To find the extrema, you have to look at the values where $H^\prime=0$ then use the second derivative test and so on to classify them.

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$H´(t) = 6 t⁴ \cdot t \cdot (1-t)(1+t)$. Then we can see that $ H´(t) >0 $ in $ (- \infty,-1) \cup (0,1) $ and $ H´(t)<0 $ in $ (-1,0) \cup(1,+\infty) $. Thus $ H $ increases in $ (- \infty,-1) \cup (0,1) $, decreases in $ (-1,0) \cup(1,+\infty) $ and possui maximum in $ H(1) = H(-1) = 1/4 $. Finally, how $ \lim_{t\rightarrow + \infty}H(t)= -\infty$. $ H $ does not have minimum.