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?
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?
$H^{\prime}(t)=6t^5-8\cdot \frac{3}{4}t^7 = 6t^5 - 6t^7 = 6t^5(1-t^2) = 6t^5(1-t)(1+t).$
When is $H^{\prime}(t) = 0$?
On what intervals is the function increasing? 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.
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!!)
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.
$H´(t) = 6 t⁴ \cdot t \cdot (1-t)(1+t)$. 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.