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