Question

HELP! WILL 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.

Answer 1

So the question is asking for the x values when g'(x) = 0?

Answer 2

I think it just want the critical points, but I need also to figure out the original formula.

Answer 3

The critical points are when the derivative equals 0

Answer 4

I found the critical points to be 4 and -4. I forgot to put the rest of the questions to the problem on here.

2. 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.
3. On what intervals is the graph of g concave down? Justify your answer.
4. Write an equation for the tangent line to the graph of g at the point where x = 3.
5. Does this tangent line lie above or below the graph at this point? Justify your answer

Answer 5

Aha, so doing these steps will result in the original function ultimately

Answer 6

For point 2, you need to check if it's going up or down on both sides of the critical points

Answer 7

How do I do that @aleroth

Answer 8

Hints:
1. how to find g(x)
g'(x)=(x^2-16)/(x-2)
g(x)=\(\int g'(x)dx\)
and the constant of integration can be found using the relation
g(3)=4

2. Integration
To help integrate, break g'(x) into partial fractions:
\(g'(x)=(x^2-16)/(x-2)=x-\frac{12}{x-2}\)+2
Once g(x) is found, the remaining questions are straightforward.
On examining g(x), the statement appears flawed if g(x) is real:
"Let g be a function that is defined for \(all\) x, x ≠ 2,... "

3. critical points are relevant only when g(x) is defined, i.e. when x \(\in\) dom g(x). Critical \(values\) are the values of the function g(x) at the critical points. So finding g(x) is important.

Answer 9

@mathmate so is the x-12/x-2+2 my g(x).

Answer 10

g'(x)=\(\dfrac{x2−16}{x−2}=x−\dfrac{12}{x−2}+2\)

We need to integrate term by term to get:
\(\int g'(x)dx =\int (x−\frac{12}{x−2}+2)dx = \dfrac{x^2+4x}{2}+\int \frac{12}{x-2}\)
\(= \dfrac{x^2+4x}{2}+12~\log(x-2)+C\)
which means that dom g(x) = \((2,+\infty \))
Note: I write log to mean \(\log_e\), and \(\log_{10}\) for log to base 10.

The rest should follow with the hints above.
If you need more help, just post.

Answer 11

@mathmate I am still sort of confused by this problem. Can we just start from the begining?

Answer 12

To find the critical numbers just solve g'(x)=0 and solve g'(x) dne . The x values of these equations that are in the domain of the original function
are the critical numbers.

Answer 13

And actually it says x=2 is not in the domain of the original function.
So you really only need to solve g'(x)=0

Answer 14

that is solve x^2-16=0 and you are done
these will be the x values where you have critical values

Answer 15

your question does say to find the critical values

Answer 16

so we don't need to find the original function

Answer 17

@freckles
I believe we need to find g(x) in order to know its domain, here's why.

1. Solution of g'(x)=0 gives x=\(\pm4\), while g(x) has a domain of \(x\in(2,\infty)\), which means that x=-4 is not a critical point and must be eliminated with a knowledge of g(x).
2. The critical value g(4) is required as part of the answer, so g(x) is required to find that.

... unless there is a different way to do that.


@krissy3039
If all this is confusing to you, we can start over step by step, when we have a time both online. Feel free to tag me when you get online.

Answer 18

@mathmate I don't see it saying anything about finding critical values.
"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."

Answer 19

also x has to be greater than 2?

Answer 20

I thought the domain was (-inf,2) union (2,inf)

Answer 21

oh and i didn't see those other questions

Answer 22

"Find all values of x where the graph of g has a critical value."
You're quite right! I misread the question. We just need the values of x, not the critical-values, mea culpa!

However, we still need g(x) to find the domain that contains the critical points.

Answer 23

As I mentioned earlier, I do admit that the question is misleading in saying that
"Let g be a function that is defined for all x, x ≠ 2,...". You may have included -4 as a critical point because of the statement.
This statement contradicts with the actual function g(x).

Answer 24

\[\int\limits_{}^{}\frac{1}{x} dx=\ln|x|+C\]

Answer 25

\[\int\limits_{}^{}\frac{1}{x-2}=\ln|x-2|+C\]

Answer 26

@krissy3039
When you get online, you can tag me or @freckles . We're on the same wavelength, so no worries.