Find the intervals of inc and dec & local max and min of h(x)=x^3/2-256x? using 1st derivative test.?
thanks...this one is really getting me!!!
First find $\displaystyle h'(x)$:
$\displaystyle h'(x)=\tfrac{3}{2}x^{\frac{1}{2}}-256$
Now set this equal to zero:
$\displaystyle \tfrac{3}{2}x^{\frac{1}{2}}-256=0\implies x^{\frac{1}{2}}=\tfrac{512}{3}\implies x=\tfrac{262144}{9}\approx 29127.1$
This is the only critical point. So we investigate the behavior of $\displaystyle h'(x)$ over the intervals $\displaystyle \left(0,\tfrac{262144}{9}\right)\text{ and }\left(\tfrac{262144}{9},\infty\right)$
Pick a point in these intervals and plug them into $\displaystyle h'(x)$, for simplicity, I'll pick 1 and 360,000 (the last one seems large, but when we take the square root of it, we get a nice number to work with...no decimals involved).
$\displaystyle h'(1)=\tfrac{3}{2}(1)^{\frac{1}{2}}-256=\color{red}\boxed{-254.5}$
It is decreasing here.
$\displaystyle h'(360,000)=\tfrac{3}{2}(360,000)^{\frac{1}{2}}-256=\color{red}\boxed{644}$
It is increasing here.
So to answer the first part of the question, the interval where its:
increasing -- $\displaystyle \left(\tfrac{262144}{9},\infty\right)$
decreasing -- $\displaystyle \left(0,\tfrac{262144}{9}\right)$
where $\displaystyle \tfrac{262144}{9}\approx 29127.111$
Now, take note that a maximum occurs when a function goes from increasing to decreasing at a critical point, and a minimum occurs when a function goes from decreasing to increasing at a critical point. Which one do you think it is? What is the coordinates of the maximum/minimum? I leave that for you to do.
I hope this makes sense!
--Chris