Stationary points using table of signs for f' = (x^2 +2x)/(x+1)

[wduk]

Hello

I am doing a table of signs for f'(x) which is (x^2 +2x)/(x+1)^2 which had stationary points (-2,-4) and (0,0).

So my table has:


x^2 + 2x and (x+1)^2

Now the first interval i am using is (-inf, -2) but for x^2 + 2x it can be both negative and positive. For example, -1.9 results in negative, -15 results in positive.

So i am a bit confused how a table of signs is going to help me here since i can get two different results - does the table of signs not always work in some situations?

EDIT: oops, i'm stupid! -1.9 is not even in the interval ! Never mind it works fine now

 

[mmm4444bot]

I am doing a table of signs for f'(x) which is (x^2 +2x)/(x+1) which had stationary points (-2,-4) and (0,0).

So my table has:

x^2 + 2x and (x+1)^2

Now the first interval i am using is (-inf, -2) but for x^2 + 2x it can be both negative and positive. For example, -1.9 results in negative, -15 results in positive.

That value lies within your first interval; you need to use only values that lie within the interval that you're currently testing.

 

[HallsofIvy]

\(\displaystyle x^2+ x= x(x+ 1)\). If x< -1, both factors are negative so the product is positive. If -1< x< 0, x is negative but x+1 is positive. The product is negative. If 0< x, both factors are positive so the product is positive.

Why do you mention \(\displaystyle (x+ 1)^2\)? In your problem, the denominator is just x+ 1.

Added: If the fraction is, in fact \(\displaystyle \frac{x^2+ x}{x+ 1}\), that is the same as \(\displaystyle \frac{x(x+ 1)}{x+ 1}\) which is undefined at x= -1 and equal to x for all other x. That is equal to 0 only at x= 0, negative for x< 0 (other than x= -1), and positive for x>0.

If it is \(\displaystyle \frac{x^2+ x}{(x+ 1)^2}\), that is the same as \(\displaystyle \frac{x(x+1)}{(x+ 1)^2}\) which is undefined at x= -1 and is equal to \(\displaystyle \frac{x}{x+1}\) for all x. That is 0 for x= 0, positive for x< -1, undefined for x= -1, negative for -1< x< 0, and positive for x> 0.

 

[wduk]

\(\displaystyle x^2+ x= x(x+ 1)\). If x< -1, both factors are negative so the product is positive. If -1< x< 0, x is negative but x+1 is positive. The product is negative. If 0< x, both factors are positive so the product is positive.

Why do you mention \(\displaystyle (x+ 1)^2\)? In your problem, the denominator is just x+ 1.

Added: If the fraction is, in fact \(\displaystyle \frac{x^2+ x}{x+ 1}\), that is the same as \(\displaystyle \frac{x(x+ 1)}{x+ 1}\) which is undefined at x= -1 and equal to x for all other x. That is equal to 0 only at x= 0, negative for x< 0 (other than x= -1), and positive for x>0.

If it is \(\displaystyle \frac{x^2+ x}{(x+ 1)^2}\), that is the same as \(\displaystyle \frac{x(x+1)}{(x+ 1)^2}\) which is undefined at x= -1 and is equal to \(\displaystyle \frac{x}{x+1}\) for all x. That is 0 for x= 0, positive for x< -1, undefined for x= -1, negative for -1< x< 0, and positive for x> 0.

Sorry that was a typo i didn't realise, the denominator is definitely squared, not sure why i didn't write it.