Problem

(a) Find the intervals of increase or decrease. …

04:15

Problem 42 Hard Difficulty

(a) Find the intervals of increase or decrease.
(b) Find the local maximum and minimum values.
(c) Find the intervals of concavity and the inflection points.
(d) Use the information from parts $ (a) - (c) $ to sketch the graph.
Check your work with a graphing device if you have one.

$ h(x) = 5x^3 - 3x^5 $

Answer

a) $h(x)$ is increasing on $(-1,0) \cup(0,1)$
$h(x)$ is decreasing on $(-\infty,-1) \cup(1, \infty)$
b) Local minima at $x=-1$ and maxima at $x=1$
c) The inflection points are where $h$ changes its direction of concavity:
$\left(-\frac{1}{\sqrt{2},-1.237 ),}(0,0),\left(\frac{1}{\sqrt{2}}, 1.237\right)\right.$
d) SEE GRAPH

Topics

Derivatives

Differentiation

Volume

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Discussion

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Video Transcript

he has close the would name right here. So we're gonna first find the derivative of H. We got 15 x square minus 15 X to the fourth. Then we take negative 15 up square. Get X minus one X plus one. So it's increasing. Negative one calm. A zero zero comma one and decreasing. Negative. Infinity. Negative one, one comma. Infinity Report. Be so we're gonna see that that X is equal to negative one. It goes from decreasing to increasing, so it's a minimum and X equals one. It goes from increasing to decreasing, so it's a maximum. Now we're gonna find the second derivative, which is equal to 30 X minus 60 x cubed. We're gonna find where it's equal to zero and we get X is equal to zero. X is equal to plus or minus one over square root of two. So we're gonna first look up the interval. Negative. Infinity. Pullman Negative One over square root of two. We choose negative one. And what bigger than zero Since h of negative one is bigger than zero, it's actually 30. We go negative one over square root of two comma zero. We look at the second derivative of negative 0.1. This is less than zero. You look from zero comma, one over square root of two, and we choose 0.1, which is bigger than zero. And finally we look at one over square root of two to infinity, and we look at one which is less than zero. We clung it in to the regular equation when yet negative 1.237 for each of 00 for H of one over square root of two. That is around 1.237 So we get inflection points to be negative. One over square root of two common negative 1.237 zero comma zero and one of her skirt of two comma 1.237 Next, we're gonna draw a quick for us, and it looks something like this

Problem