# Need to find the intervals where the function increase, decrease & concavity

but why? i simplified already

Check your work again.

if
f"(x)=(-2x+(x^(2)+9)(4x^(3)-36x))/(x^(2)-9)^2

(4x^(3)-36x) = 4x(x^2 -9)

f"(x)= 4x(-2x+(x^(2)+9)/(x^(2)-9)

Which I highly doubt considering the numerator is of a higher degree than the denominator.

You answer should look like
k(x)( ax^2 + b) / (x^2 -9)^3 for some integers k,a,b.

ok without that thing i did, the derivative is like this:
f"(x)=(-2x(x^(2)-9)^2)+(x^(2)-9)(4x^(3)-36x)/(x^(2)-9)^4

Mark44
Mentor
Edit: fixed exponent.
Unsimplified, the second derivative is
$$f''(x) = \frac{(x^2 - 9)^2(-2x) + (x^2 + 9)(4x(x^2 - 9))}{(x^2 - 9)^4}$$

This can be simplified somewhat by removing common factors of the numerator and denominator.

I can see that you are being diligent about putting parentheses in, but there are some you use that are not needed. Any time an exponent is a positive integer, you don't need parentheses around it. Doing so just makes what you have written harder to read and understand.

For example, there is no reason to write x^(2). x^2 means the same thing and is clearer. On the other hand, you need parentheses for x^(-1) and e^(2x).

Last edited:
must ask, when the second derivative:
the second part of the term (-2x)(x^2-9) doesn't get squared like this:
"(-2x)(x^2-9)^2"?

Mark44
Mentor
Right, that factor in the first term should be (x^2 - 9)^2. I had it in my notes, but neglected to write it in my post. I've gone back and edited my post.

When you simplify the numerator, resist the urge to multiply everything out. The best way to do things is to find write the numerator in factored form. That way it will be easier to determine the sign of f''(x).

so... what is left now is to try to simplify and evaluate the second derivative right?

if I'm not mistaken, after simplifying, the derivative ends like this:
f"(x)=(2x(x^2 + 27))/(x^2 - 9)^3

Last edited:
Mark44
Mentor
Yes, that looks fine.

yay! now i can sketch the graph!