Help with calculus find where it's increasing/decreasing Point of inflection, etc...

[zumba]

I would greatly appreciate some help with these 3 problems, and if you could, explain them in as much detail possible.

Problem 1.

For F(x) =( (x^2)-4)/((x^2)-2)

1. Identify the domain of f and any symmetries the curve may have.
2. Find the derivatves f' and f"
3.Find the critical points of f, if any, and identify the function's behavior at each one.
4. Find where the curve is increasing and where it is decreasing.
5.Find the points of inflection, if any occur, and determine the concavity of the curve.
6.Identify any asymptotes that may exist.
7. Plot key points, such as intercepts and points found in steps 3-5 and sketch the curve together with any asymptotes that exist.


Problem 2. and 3.

For the following functions find the critical points of f if any, and indentify the intervals where it's increasing/decreasing by the x,y coordinates.

2. (x^2)+(2/x)

3. (x^3)-(4x)+(6)

Once again, thanks for any help with these problems.

 

[mmm4444bot]

What have you thought about or tried on each of these, thus far? In other words, where are you stuck specifically?

In the first exercise, do you understand the meaning of the word "domain"?

 

[zumba]

Well Honestly, I do not think I am doing the problem correctly. Once I find the first derivative (4x/(x^2-2)^2) I set it equal to zero, and I get a critical point of x = 1. I do not believe that is correct though. Then, moving on, I can’t figure out the second derivative. I have done plenty of these problems; it is just this specific one that is giving me trouble. Truthfully speaking, I just need someone to break it down for me so I have a better understanding. The second and third problem after finding the first derivative, I am having trouble solving for x to find the critical points as well. To answer your question, yes I do know the meaning of the word domain.

 

[pappus]

Well Honestly, I do not think I am doing the problem correctly. Once I find the first derivative (4x/(x^2-2)^2) <-- OK!
I set it equal to zero, and I get a critical point of x = 1. <-- plug in this value into the term of the derivative: \(\displaystyle \displaystyle{\frac{4 \cdot 1}{(1^2-2)^2}\ne 0 }\)
I do not believe that is correct though. <--- correct!
Then, moving on, I can’t figure out the second derivative. I have done plenty of these problems; it is just this specific one that is giving me trouble. Truthfully speaking, I just need someone to break it down for me so I have a better understanding. The second and third problem after finding the first derivative, I am having trouble solving for x to find the critical points as well. To answer your question, yes I do know the meaning of the word domain.

To determine the 2nd derivative use quotient rule and chain rule again:

\(\displaystyle \displaystyle{f''(x)=\frac{(x^2-2)^2 \cdot 4 - 4x \cdot 2(x^2-2) \cdot 2x}{(x^2-2)^4}}\)

Don't expand the numerator but factor out \(\displaystyle (x^2-2)\) and cancel. (You are allowed to cancel because the domain excludes the case that \(\displaystyle x^2-2 = 0\) )

 

[zumba]

To determine the 2nd derivative use quotient rule and chain rule again:

\(\displaystyle \displaystyle{f''(x)=\frac{(x^2-2)^2 \cdot 4 - 4x \cdot 2(x^2-2) \cdot 2x}{(x^2-2)^4}}\)

Don't expand the numerator but factor out \(\displaystyle (x^2-2)\) and cancel. (You are allowed to cancel because the domain excludes the case that \(\displaystyle x^2-2 = 0\) )

Ah, I see now. Thanks for the help. I would really appreciate some more guidance on number two, finding the critical points. f'(x) = (2x-2)/(x^2) I know now that I set it equal to zero and solve for x. Although, I'm a little confused. Do I set the numerator and denominator both equal to zero seperately and solve for x, or solve the whole equation for x?

 

[mmm4444bot]

f'(x) = (2x-2)/(x^2)

I know now that I set it equal to zero and solve for x.

Do I set the numerator and denominator both equal to zero seperately and solve for x

No, you do not look for values of x that make the denominator zero because those value(s) are not in a function's domain.

In fact, no fraction may have zero in the denominator because division by zero is not defined in the Real number system.

By the way, your derivative above is not correct.