Question

FAN +MEDAL
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.

Answer 1

1. I got the critical values of x= 4 , -4
2. I got that since g''(x) >0 for x=4 , -4 that they are both local minimums.
3. I am stuck on finding the intervals of concave down.

Answer 2

@Astrophysics Can you help?

Answer 3

@mathmale ?

Answer 4

@ParthKohli

Answer 5

"x ≠ 2" ?? Why the focus on x=2, if the function is defined for all x?

Answer 6

This statment, "g′(x) = x^2-16/x-2 with x ≠ 2," is unusual. Why "with x=2?"

Answer 7

x cannot equal 2. Theres a vertical asymtote there.

Answer 8

the important thing here is that you're given the derivative of function g(x).
How do you go about using this derivative to determine the critical values of g(x)?

Answer 9

OK. then write \[x \neq2\]

Answer 10

Ive already figure out the critical values. Its 4, -4

Answer 11

This is the opposite of x=2, which was the reason I needed to ask for clarification.

Answer 12

Please demonstrate that 4 and -4 are indeed critical values of g(x). How would you do that?

Answer 13

You set the derivative equal to 0. then solve for x. leading you to obviously get 4, -4

Answer 14

So, if you substitute x=4 into g'(x), you should obtain what result?

Answer 15

You would get 0.

Answer 16

Great. Let's move on.

Answer 17

What do you need help with?

Answer 18

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

Answer 19

Starting with the given g '(x), find the second derivative, g ''(x).

Answer 20

\[g''(x)= \frac{ x ^{2}-4x+16 }{ (x-2)^{2} }\]

Answer 21

I'm not going to check that. Given that you started with a rational function, your obtaining a rational function as the second derivative is approrpriate. Please set this second derivative = to 0 and solve for x. This is to set up intervals on which g(x) is concave up and on which it's concave down.

Answer 22

Hints:

If you set this g ''(x)=0 and want to solve for x, that's equivalent to setting the numerator = 0 and solving for x.

Note that g ''(x) is not definied at x=2.

Plot x=2 on a number line.

If the roots of g ''(x) are real, also graph them on the same number line.

Answer 23

Thats the part I dont understand.

Answer 24

2+2i sqrt3 ? like i dont understand

Answer 25

Set the numerator of this rational function = to 0. In other words,\[x^2-4x+16=0\]
I see you've already done this and have solved for the roots. Your roots are "complex". No point in graphing them on the real x axis.

Answer 26

the only relevant real number to graph on the x-axis is x=2.

Choose an x value smaller than 2 and then another one greater than 2.
Substitute these 2 numbers into the 2nd derivative, only for the purpose of determining the sign of the 2nd derivative at each value of x.

Results?

Answer 27

I chose 3 and 1. Both god positive 13

Answer 28

On the interval\[(-\infty,2),\] the second derivative is positive? negative?
On the interval \[(2m \infty), the second der. is positive? \neg?\]

Answer 29

So, your 2nd derivative is + on both sides of x=2. What does that tell you about the direction of concavity of each half of the graph?

Answer 30

Its concave up.

Answer 31

which means since the second derivative is always positive, it is never concave down, correct?

Answer 32

yes. very good. However, I'd prefer you mention that x=2. the sec. deriv. is undefinited there, so I wouldn't want to say the curve of g(x) is concave up at x=2. Instead, use sets, as I did above.

Answer 33

Okay! Awesome. For the next step, Write an equation for the tangent line to the graph of g at the point where x = 3. Would it be:
y-y1=m(x-x1)
y-g(3) = g'(3)(x-3)
y-4=-7(x-3)

Answer 34

Yes, very good. You already have a formula for the slope of the t. l, and need only subst. x=3 into it to determine the slope of the tan. line at x=3. How do you propose to find the value of g(3)?

Answer 35

g(3) = 4 with the given information that the problem gave us. Then to find g'(3) I substituted 3 for x in g'(x) and got -7.

Answer 36

I haven't attempted that. But you demonstrate considerable competence, so I'll go along with your result here.

Answer 37

so plugging it all into the equation, so far i got y-4 = -7 (x-3)

Answer 38

Again, that strikes me as appropriate , although I haven't done the actual calculations myself. Nice going on your part!

Answer 39

Would I need to get it into y=mx+ form? or would that be the equation for the tangent line at x=3?

Answer 40

hello?

Answer 41

that's be an appropriate form, the simplest you could choose. Sorry for the delay; I got sidetracked. Are you comfortable with the solution you've reached?

Answer 42

Yes, completely perfect, Thank you!