Question

Consider the following function f(x)=x^2/[x^2-9]

f(x) is increasing on the interval(s)

f(x) is decreasing on the interval(s)

f(x) has 2 vertical asymptotes x=

f(x) is concave up on the interval(s)

f(x) concave down on the interval(s)

I've been stuck on these parts for about an hour. I would really like some assistance.

Answer 1

when f'(x) > 0 it's increasing
when f'(x) < 0 it's decreasing
vertical asymptotes: put the denominator equal to zero and solve for x.
So when x^2 - 9 = 0
(Just an extra bit of info: for horizontal asymptotes divide by the highest power of x and then apply the limit as x -> infinity)
Concave up - It's when f''(x) < 0 I think
Concave down - It's when f''(x) > 0 I think

Answer 2

Can you please explain what you mean by when f'(x) > 0. I know you should be able to just graph it and see when it's increasing and decreasing, but for some reason I cannot get the correct answer.

Can you please explain increasing and decreasing, concave up and down with more details