Problem

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

04:36

Problem 37 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.

$ f(x) = x^3 - 12x + 2 $

Answer

(a) $f(x)=x^{3}-12 x+2 \Rightarrow f^{\prime}(x)=3 x^{2}-12=3\left(x^{2}-4\right)=3(x+2)(x-2)$. $f^{\prime}(x)>0 \Leftrightarrow x<-2$ or $x>2$
and $f^{\prime}(x)<0 \Leftrightarrow-2<x<2 .$ So $f$ is increasing on $(-\infty,-2)$ and $(2, \infty)$ and $f$ is decreasing on (-2,2)
(b) $f$ changes from increasing to decreasing at $x=-2,$ so $f(-2)=18$ is a local maximum value. $f$ changes from decreasing
to increasing at $x=2,$ so $f(2)=-14$ is a local minimum value.
(c) $f^{\prime \prime}(x)=6 x . \quad f^{\prime \prime}(x)=0 \Leftrightarrow x=0 . f^{\prime \prime}(x)>0$ on $(0, \infty)$ and
$f^{\prime \prime}(x)<0$ on $(-\infty, 0) .$ So $f$ is concave upward on $(0, \infty)$ and $f$ is
concave downward on $(-\infty, 0) .$ There is an inflection point at (0,2)

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 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

Video Transcript

So in part a we're going to find where our function F of X is increasing and decreasing. So I've just written down the derivative already and I just use the power rule to find this. And so we're just going to figure out where are derivative is positive and where it's negative. So the easiest way to do this I think is to find where it's actually equal to zero first. Since then we can know what intervals to look at to see how our the sign of our derivative F. Prime of X. And so I'm just solving for X. Here we get four is equal to X squared, so X is equal to plus or minus two. So a plus or minus two. We have f prime of X equaling zero. And now if we look at the intervals X being less than negative two, X being between negative two and two, and X being greater than X being greater than two. Sorry. And we look at the sign of our derivative, we'll know where we're increasing and decreasing. So if we plug in an X value less than negative two, let's just say negative three. We get nine times three minus 12 which is positive. So are derivative. F prime of X is greater than zero. We plug in a value between negative two and two. Let's just say zero. We get zero minus 12 which is negative. So are derivative F prime of X is less than zero. And then if we plug in the value of X greater than two, say three, we get nine times three again, 27 minus 12 is positive. So the intervals in which were increasing, we're going to be the first and last intervals. So from negative infinity to negative two and from two to infinity. And then we're decreasing in this other interval from negative to 22 Now, if we go to part B, we're going to do is find local maximum and minimum values. And we already found the zeros of our derivative up here at plus and minus two. So we're just going to look at actually, we're just going to look at the information we already found. So you want to look at values of X less than negative two. We can see that we're going positive and the values after negative to our negative. So we're going from increasing slope to decreasing slope at X is equal to negative two, which means we have a maximum value. So we have a local max at X is equal to negative two. And if we look at the values before too, but greater than negative two were negative and then greater than two were positive. So we're going from negative values to positive values are decreasing slope to increasing slope, which means it is a local minimum. And now for part, see what we're gonna do is find the intervals in which our function f of X is concave up in concave down. And then we're also going to find the intervals are the points of inflection. So I've written down this first derivative because we're going to use it to find our second derivative, which then we can use to figure out con cavity. So I'm just gonna use the power rule again, find the second derivative, bring down the exponents gets six X. And this 12 goes to zero. And now this one we just want to look at where this is negative and where it's positive. Well you can see that's gonna be negative whenever X is negative, it's gonna be positive whenever X is positive. So we're concave up from zero to infinity. Since we need positive values are second derivative to be concave up, Merkel cave down from negative infinity to zero. And then we also have an inflection point at zero, since that's where our second derivative is equal to zero. And we know we're going from negative to positive. So we're actually crossing that X axis at X is equal to zero, so inflection point at X is equal to zero. And if we plug that back into our original equation this is actually equal to two. So we have the inflection point is at this 0.0.0 comma two. And now for part D, what we're gonna do is use all this information that we found and actually try and sketch a graph of of our functions. So we know that F zero is two, so zero comma two. And that's an inflection point, we know that F of zero is, oh never mind. We know that. We know the F. Of zero is too. So I plugged that point in here and now we want to look at, well we're concave up before we're sorry we're concave down when we're negative. So I'm just gonna put concave down, concave up. So I remember and then we have a local max at X is equal to negative two. A local men X is equal to two. So um what we're gonna do now is I'm actually going to just plug in F of two into our equation which was X the third minus 12 X two to the third minus 12 times two plus two. This is equal to eight minus 24 plus two. Just equal to negative 14. So we have this point at to negative 14. I'm just gonna put down here I'm not going to get super specific and now I'm gonna figure out f of negative two. So the only difference here is we're gonna have plus 24 instead of minus and this is going to be eight plus 24 32 34. So you have a point F is equal to negative two 34. Just stay up here and now we can look at and this is at negative two and we know that this is a maximum value. This is a minimum value. Mhm. And now we can look at we're increasing from negative infinity to negative two and we are concave down right, so we need to be increasing in concave down and then we're decreasing in concave actually instead of so we're increasing in concave down. And then at this point here too we start to be concave up since that's an inflection point and we are now still decreasing until we get to this point at two comma negative 14 and then we start increasing again. So we need to go all the way down into this point to and we can start increasing again and actually shouldn't make it look like we start to go concave down again since we don't have another inflection point. So this is somewhat of what our graph should look like. Again, we have these local max is that x is equal to negative two. Local men addicts is equal to two. We're decreasing from negative to to to or increasing from negative infinity to negative two and from two to infinity. And we're concave up where we need to be re concave down where we need to be. We have this inflection point at zero comma two. So yes, this is indeed a good rough sketch of our graph.

Kian M.

Oregon State University

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

Problem