Problem

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

04:17

Problem 39 Medium 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.

$ f(x) = \frac{1}{2} x^4 - 4x^2 + 3 $

Answer

(a) $f(x)=\frac{1}{2} x^{4}-4 x^{2}+3 \Rightarrow f^{\prime}(x)=2 x^{3}-8 x=2 x\left(x^{2}-4\right)=2 x(x+2)(x-2) . \quad f^{\prime}(x)>0 \Leftrightarrow-2<x<0$
or $x>2,$ and $f^{\prime}(x)<0 \Leftrightarrow x<-2$ or $0<x<2 .$ So $f$ is increasing on (-2,0) and $(2, \infty)$ and $f$ is decreasing on
$(-\infty,-2)$ and (0,2)
(b) $f$ changes from increasing to decreasing at $x=0,$ so $f(0)=3$ is a local maximum value.
$f$ changes from decreasing to increasing at $x=\pm 2,$ so $f(\pm 2)=-5$ is a local minimum value.
(c) $f^{\prime \prime}(x)=6 x^{2}-8=6\left(x^{2}-\frac{4}{3}\right)=6\left(x+\frac{2}{\sqrt{3}}\right)\left(x-\frac{2}{\sqrt{3}}\right)$
$f^{\prime \prime}(x)=0 \Leftrightarrow x=\pm \frac{2}{\sqrt{3}}, \quad f^{\prime \prime}(x)>0$ on $\left(-\infty,-\frac{2}{\sqrt{3}}\right)$ and $\left(\frac{2}{\sqrt{3}}, \infty\right)$
and $f^{\prime \prime}(x)<0$ on $\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right) \cdot$ So $f$ is $\mathrm{CU}$ on $\left(-\infty,-\frac{2}{\sqrt{3}}\right)$ and
$\left(\frac{2}{\sqrt{3}}, \infty\right),$ and $f$ is $\mathrm{CD}$ on $\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right) .$ There are inflection points at
$\left(\pm \frac{2}{\sqrt{3}},-\frac{13}{9}\right)$

More Answers

01:43

Carson M.

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Related Topics

Derivatives

Differentiation

Volume

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Kayleah T.

Harvey Mudd College

Joseph L.

Boston College

Watch More Solved Questions in Chapter 4

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 93

Video Transcript

and this problem, we are learning how derivatives affect the shape of a graph. More specifically, this problem is asking us how we can use the derivative to deduce important information about the function, like where the function is increasing or decreasing three intervals of con cavity. And it's critical numbers. And this is going to become important later on calculus when you start to learn curve sketching, so getting a great understanding of this now is going to be very helpful for you. Sorry. Pardon me for part A were given the function F of X equals in one half X to the fourth minus four X squared plus three. So we need to find the critical numbers and the intervals of con cavity. So the first thing that we're going to do is take the derivative F prime of X equals two X cubed minus eight x We could do some. Factoring will get two x times X minus two times X plus two. We're going to set that equal to zero to get the critical numbers. Then we'll solve for X so x zero X s two and excess negative too. So then what? We're going to Dio is we're going to create a table. You don't have to make a table, but just keep track of the intervals. We're going to make sure that we have the intervals using the critical numbers and then determine if our function would be increasing or decreasing in that interval. So the first interval is negative. Infinity too negative too. And our function is decreasing there. And if you don't know how I'm getting that, all you have to do is take a number within that interval, plug it into the function and see if you get a positive or a negative. For the second interval, we'd have negative 2 to 0 and we see our function is increasing there. Our third interval interval. Pardon me is from 0 to 2 and our function is decreasing there. And then finally, our last interval is to to infinity and our function is increasing there. So this is just another way to right. We found on the table at the vex increasing when X is an interval. Negative 20 and two comma infinity and f of X is decreasing on the interval when X is in thes two intervals negative and they needed to negative negative two or two or pardon me, you 02 and then for party were told. What are the maxes and men's of this function? What we have the critical numbers. All we have to do now is plugged them back into our original function. So f of negative two is negative. Five f of two was negative. Five and F of zero is three. So clearly we have a minimum, a negative two and two and a maximum at zero for part C, we're told. Let's check the con cavity. So when you hear the word con cavity your mind, you immediately go to second derivative. We need the second derivative test to determine the intervals of con Cavity. So F double prime of X is six x squared minus eight. And now this one's a little bit more confusing than before. We have to do some clever factoring. So this is the same thing as saying six times X squared, minus 4/3. And then we want critical numbers, Remember? So we have to factor this further. We're going to get six times X minus two over the square root of three times X plus two over the square root of three. So then we have our critical numbers so we can determine con cavity. Our function is con cave up on the interval. Negative infinity negative to over screwed of three and two over. Screwed of three common infinity And then we see a con cave down behavior on the interval negative to over the square root of 3 to 2 over the square root of three. So what does this mean? We change from going from concave up to con cave down So we see an inflection point when x is negative two over the skirt of three and when X is two over the square root of three. And now finally the problem asks us, Well, what's going to happen with the graph? Can we check our information from the graph? This is the graph of the function and you can clearly see that our information that we found using just differential calculus matches the function. We see our minimums where we found them, our maximum where we found them. Our con cavity matches as well. So I hope this problem helped to understand a little bit more about how the derivative, um, affect the shape of the graph more specifically, how we can use the derivative to deduce important information from a function such as the intervals of increase in decrease and it's common cavity.

Madi S.

University of Denver

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Related Topics

Derivatives

Differentiation

Volume

Top Calculus 2 / BC Educators

Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Kayleah T.

Harvey Mudd College

Joseph L.

Boston College