Question

PLEASE HELP
Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is g′(x)=${x^2–16}/{x−2}$, with x ≠ 2.

Find all values of x where the graph of g has a critical value.

For each critical value, state whether the graph of g has a local maximum, local minimum or neither. You must justify your answers with a complete sentence.

On what intervals is the graph of g concave down? Justify your answer.

Write an equation for the tangent line to the graph of g at the point where x = 3.

Does this tangent line lie above or below the graph at th

Answer 1

@3mar

Answer 2

@mathstudent55

Answer 3

@SolomonZelman

Answer 4

critical values are values of x when f'(x) =0

Answer 5

Let \(\color{black}{\displaystyle g }\) be a function that is defined for all \(\color{black}{\displaystyle x }\), \(\color{black}{\displaystyle x\ne 2}\), such that \(\color{black}{\displaystyle g(3)=4 }\) and
the derivative of \(\color{black}{\displaystyle g }\) is \(\color{black}{\displaystyle g'(x)=(x^2-16)/(x-2) }\), with \(\color{black}{\displaystyle x\ne 2}\).

Answer 6

Critical points, in general, occur in 3 different cases/conditions.

1. The function is restricted to an interval. Then, any closed boundary is a critical point - (an open boundary is not part of the function so obviously it's not a critical point for the function.)

2. Any value(s) of x for which the derivative of the function is undefined, provided that the function is defined on this value.

3. All values of x that satisfy f'(x)=0 are critical numbers.

Answer 7

(It has to meet any of the 3 conditions to qualify for critical point.)

Answer 8

so we need to look for when the function is undefined and when it equals 0?

Answer 9

Well, as you can note, #1 does not apply because no interval is given (and certainly not any with closed boundaries), and so won't #2 (because there is only one value where f' is undefined, at x=2, for which not only the derivative, but also the function is undefined.

Answer 10

So, the only critical numbers you will have are the solutions to \(\color{black}{\displaystyle f'(x)=0 }\).

Answer 11

Can you set the derivative equal to 0, and solve for x, please?

Answer 12

\(\color{black}{\displaystyle f'(x)=0 }\)

\(\color{black}{\displaystyle \frac{x^2-16}{x-2}=0 }\)

this is what you got to solve for x, can you?

Answer 13

would that just equal x^2-16= 0, so x^2=16..... x=4

Answer 14

yes, but there is another solution besides x=4.

Answer 15

would it be x=-4

Answer 16

yes.

Answer 17

so a critical value at x=4 and one at x=-4. is there another step to find another critical value

Answer 18

\(\color{black}{\displaystyle \frac{x^2-16}{x-2}=0 }\)

\(\color{black}{\displaystyle \frac{x^2-16}{x-2}\color{red}{\times (x-2)}=0\color{red}{\times (x-2)} }\)

\(\color{black}{\displaystyle x^2-16=0 }\)

\(\color{black}{\displaystyle x^2=16 }\)

\(\color{black}{\displaystyle \sqrt{x^2}=\sqrt{16} }\)

\(\color{black}{\displaystyle |x|=4 }\)

\(\color{black}{\displaystyle x=4,~-4. }\)

Answer 19

This is the full/total algebra behind this.

Answer 20

Yes, your critical numbers are x=-4 and x=4.

Answer 21

okay, that makes sense

Answer 22

Yes, basically applying the definition of absolute value:
\(\color{black}{\displaystyle \sqrt{z^2}=|z|}\) for all \(\color{black}{\displaystyle z\in \mathbb{R}}\).

Answer 23

K.

Answer 24

1. Now, we have to determine for each of the critical points (x=-4,4) whether this point is local maximum or minimum.

Answer 25

so would you graph this or what step would you take to determine that

Answer 26

We can do it mathematically, with some conceptual:)

Answer 27

Suppose for example that a function \(\color{black}{\displaystyle f(x) }\) has a critical value at \(\color{black}{\displaystyle x=a }\).

You can make a logical note that if \(\color{black}{\displaystyle f'(a-\delta) >0 }\) and \(\color{black}{\displaystyle f'(a+\delta) <0 }\)
that the function is increasing before \(\color{black}{\displaystyle x=a }\) and decreases after \(\color{black}{\displaystyle x=a}\),
which allows us to imply that \(\color{black}{\displaystyle x=a }\) is the maximum (not minimum).

Answer 28

(For sufficiently small \(\delta\), for example \(\delta = 0.1\))

Answer 29

oh ok makes sense, so how do you apply 4 and -4 into this

Answer 30

Similarly if \(\color{black}{\displaystyle f'(a-\delta) <0 }\) and \(\color{black}{\displaystyle f'(a+\delta) >0 }\), then the function is decreasing before \(\color{black}{\displaystyle x=a }\) and increases after \(\color{black}{\displaystyle x=a}\), which allows us to imply that \(\color{black}{\displaystyle x=a }\) is the minimum (not maximum).

Answer 31

In our case ....

Answer 32

\(\color{black}{\displaystyle f'(x)=\frac{x^2-16}{x-2} }\)

Answer 33

Can you tell me, near x=4, what is the slope from the left and from the right?

Answer 34

(You can choose a value like \(\color{black}{\displaystyle \delta =0.2}\), or any sufficiently small value that you may wish.)

Answer 35

\(\color{black}{\displaystyle f'(3.8)=?}\)
\(\color{black}{\displaystyle f'(4.2)=?}\)

Answer 36

would you just derive (x^2-16)/(x-2) and substitute 3.8 and 4.2

Answer 37

(You mean differentiate, not derive. The verb "derive" is not defined as taking the derivative.)

No, because (x^2-16)/(x-2) is already the derivative of your function.
(This is given in the very beginning of your problem.)

Answer 38

and as a matter of fact, you just need to tell me if they are negative or positive .
(You don't need to know the actual value.)

Answer 39

so just substitute 3.8 and 4.2 into this equation, which would get -.867 and .745.

Answer 40

i got one negative and one positive

Answer 41

Yes, very good!

Answer 42

\(\color{black}{\displaystyle f'(4-0.2)<0 }\) and \(\color{black}{\displaystyle f'(4+0.2)>0 }\).

Correct?

Answer 43

yes

Answer 44

meaning its the minimum?

Answer 45

The slope before x=4 is negative, so the function is decreasing before x=4, and then after x=4 the slope is positive so the function is increasing after x-4.

YES, x=4 is the local minimum.

Answer 46

Now, we will do the same thing for \(\color{black}{\displaystyle x=-4 }\).
(You can choose \(\delta=0.2\), or any other (positive and sufficiently small) \(\delta\).)

Answer 47

so then would we do the same thing for -4?

Answer 48

Yup yup:)

Answer 49

well before -4 it is a negative number and after it is positive

Answer 50

so would this be a local minimum as well

Answer 51

yes, you are correct again!

Answer 52

So, both points x=4 and x=-4 turn out to be minimums.

Answer 53

okay perfect!

Answer 54

(We could have tested for minimums or maximums by looking at concavity as well ... just to make a note.)

Answer 55

OK, what was the next part of the question?

Answer 56

Concavity!

Answer 57

When i graphed it it looks like the left side is concave up and the right side is concave down. the asymptote (i think thats what its called) looks like its around x=2

Answer 58

oh it has to be 2 because when you put 2 in the denominator you get 0

Answer 59

In general,
1) if \(\color{black}{\displaystyle f'(x)>0 }\), then \(\color{black}{\displaystyle f(x) }\) is increasing.
2) if \(\color{black}{\displaystyle f'(x)<0 }\), then \(\color{black}{\displaystyle f(x) }\) is decreasing.

And you can think of\(\color{black}{\displaystyle f''(x) }\) and \(\color{black}{\displaystyle f'(x) }\) as \(\color{black}{\displaystyle f''(x) }\) is the derivative of \(\color{black}{\displaystyle f'(x) }\).
So, \(\color{black}{\displaystyle f''(x) }\) is the derivative and \(\color{black}{\displaystyle f'(x) }\) is the function (if you treat \(\color{black}{\displaystyle f'(x) }\) like a function.)
That is the relationship between them.

So, same way
1) if \(\color{black}{\displaystyle f''(x)>0 }\), then \(\color{black}{\displaystyle f'(x) }\) is increasing.
2) if \(\color{black}{\displaystyle f''(x)<0 }\), then \(\color{black}{\displaystyle f'(x) }\) is decreasing.
-------------------

Making sense?

Answer 60

so concave up from -infinity to 2 and concave down form 2 to positive infinity?

Answer 61

Can we do it without graphing? Can we just solve it mathematically?

Answer 62

yes sorry

Answer 63

it's ok:)

Answer 64

Does my last long post make sense?

Answer 65

yes i think so

Answer 66

very good.

Answer 67

Concavity is basically measures the behavior of the slope of the function.

[*] The function is "concave up" iff the slope is increasing.
[*] The function is "concave down" iff the slope is decreasing.

Answer 68

Alternatively,

[*] The function is "concave up" iff \(\color{black}{\displaystyle f'(x)}\) is increasing.
[*] The function is "concave down" iff the \(\color{black}{\displaystyle f'(x)}\) is decreasing.

Answer 69

okay! so then to determine what interval it is concave down, or when f'(x) is decreasing, what would we do?

Answer 70

Normally, \(\color{black}{\displaystyle f(x)}\) is increasing iff \(\color{black}{\displaystyle f'(x)>0}\). Right? Same here!
\(\color{black}{\displaystyle f'(x)}\) is increasing iff \(\color{black}{\displaystyle f''(x)>0}\).

(The relation between f' and f, is the same as the relation between f'' and f')

Answer 71

So, to find all \(x\) values (or to find the intervals) at which the function is concave up (Note: concave up, is an equivalent way of saying \(\color{black}{\displaystyle f'(x)}\) is increasing), you need to find \(\color{black}{\displaystyle f''(x)}\) and set \(\color{black}{\displaystyle f''(x)>0}\).

Answer 72

We had:
\(\color{black}{\displaystyle f'(x)=\frac{x^2-16}{x-2}}\)

can you find \(\color{black}{\displaystyle f''(x)}\) ?

Answer 73

but we need to find it when its concave down so would we do f''(x)<0?

Answer 74

Oh, yes.

Answer 75

So, you will need to set \(\color{black}{\displaystyle f''(x)<0}\), in order to see where the slope (or the f'(x)) is decreasing.

Answer 76

What is the \(\color{black}{\displaystyle f''(x)={\tiny~}?}\)

Answer 77

so 0 = (x^2-16)/(x-2) which would simplify to x^2=16 so x= 4 or -4

Answer 78

is that correct?

Answer 79

We already set \(\color{black}{\displaystyle f'(x)=0}\) to solve for critical points.

Answer 80

You are doing the same thing again. (I don't mind to do this again, but something makes me think you don't want to redo the critical points again:))

Answer 81

oh so the derivative of that

Answer 82

yes, the derivative of f'(x), which is f''(x).

Answer 83

\(\color{black}{\displaystyle f'(x)=\frac{x^2-16}{x-2}}\)

\(\color{black}{\displaystyle f''(x)=\frac{d}{dx}\left[f'(x)\right]=\frac{d}{dx}\left[\frac{x^2-16}{x-2}\right]={\tiny~}?}\)

Answer 84

Well, if you so hate quotient rule, you can rewrite

\(\color{black}{\displaystyle f'(x)=\frac{x^2-16}{x-2}=\frac{x^2-4-12}{x-2}=x+2-\frac{12}{x-2}}\)

Answer 85

but, if you don't hate, you can differentiate just as it is.

Answer 86

x^2-4x+16/(x-2)^2

Answer 87

Yes, this is right!

Answer 88

\(\color{black}{\displaystyle f''(x)=\frac{x^2-4x+16}{(x-2)^2} }\)

Answer 89

Now set \(f''(x)<0\).

Answer 90

how do you set it equal to less than 0?

Answer 91

yes, set the second derivative less than 0.

Answer 92

\(\color{black}{\displaystyle \frac{x^2-4x+16}{(x-2)^2}<0 }\)

Answer 93

We have \(x\ne 2\) so that \((x-2)^2\ne 0\).
You can note that \((x-2)^2>0\) for all real \(x\),
so, the equivalent equation to solve would be
\(\color{black}{\displaystyle x^2-4x+16<0 }\).

Answer 94

Note also that
\(\color{black}{\displaystyle x^2-4x+16<0 }\)
\(\color{black}{\displaystyle x^2-4x+4+12<0 }\)
\(\color{black}{\displaystyle (x-2)^2+12<0 }\)

so, when will this function be concave down?

Answer 95

it will be concave down when the values of the function are greater than 2?

Answer 96

I don't think so.

Answer 97

you understand why we set to find all \(x\) such that the function is concave down,
\(\color{black}{\displaystyle \frac{x^2-4x+16}{(x-2)^2}<0 }\)

right?