Find the first derivative of f(x). Construct a sign chart for the first derivative, and find the intervals over which f(x) is increasing and decreasing.

(x-2)/(x+1)

First Derivative is: f'(x)= 3/(x+1)^2

No idea how to find the critical values from this so I can place them on my sign chart.

Question

Find the first derivative of f(x). Construct a sign chart for the first derivative, and find the intervals over which f(x) is increasing and decreasing.

(x-2)/(x+1)

First Derivative is: f'(x)= 3/(x+1)^2

No idea how to find the critical values from this so I can place them on my sign chart.

jim_thompson5910 · Answer

You need to look for the roots and the vertical asymptotes of 3/(x+1)^2

jim_thompson5910 · Answer

able to find them?

anonymous · Answer

how do I find the roots and asymptotes on this type of problem? since it's the first derivative.

jim_thompson5910 · Answer

set it equal to zero and solve for x to find the roots

3/(x+1)^2 = 0

x = ???

anonymous · Answer

none

jim_thompson5910 · Answer

good, there are no roots

jim_thompson5910 · Answer

to find the vertical asymptote(s), you set the denominator equal to zero and solve for x

anonymous · Answer

x=-1

jim_thompson5910 · Answer

so that's your only critical value

jim_thompson5910 · Answer

there's a potential (not definite) that the sign of f ' (x) could change as you pass through x = -1

anonymous · Answer

Hmm... ok I am plugging them in and I am getting -2,3  0,3  1,0.75

jim_thompson5910 · Answer

you just need two points....one point per region (on either side of x = -1)

jim_thompson5910 · Answer

but basically the outputs are all positive

which means f ' (x) > 0 for all x in the domain of f ' (x)

jim_thompson5910 · Answer

so f is increasing over its entire domain

jim_thompson5910 · Answer

no matter which point you're at on the graph of y = f(x), if you move over to the right, the point will move up (to stay on the graph)