Find the intervals on the x-axis on which the function f(x) is increasing and those which it is decreasing, where f(x) = (x)/(x^2+1)

I differentiated and got
f'(x)=(x^2+1) -(x)(2x) all over (x^2+1)^2

But I don't know what to do now...

f ' (x) = (x^2+1) -(x)(2x) all over (x^2+1)^2

= (x^2 + 1 - 2x^2)/(x^2+1)^2
= (1 - x^2)/(x^2 + 1)^2

You can be sure that the denominator will always be positive, so let' just look at the top.

recall that a function is increasing when its derivative is positive, and it is decreasing when .... negative.

so when is 1 - x^2 > 0 ??
clearly for all values between -1 and +1, that is, for all proper fractions.

so the function increases for -1 < x < 1 , and
decreases for x < -1 OR x > 1

confirmation:
http://www.wolframalpha.com/input/?i=y+%3D+%28x%29%2F%28x%5E2%2B1%29

the 2nd graph shows is better than the first.

set f' to zero, solve for x. That is either a max or min location for f(x).

example

0= x^2+1-2x^2=0(x^2+1)^2

0=-x^2+1

x=+-1

now check a point in between.
at x=0
f'(x)=1/1= positive, so from -1 to 1, x is increasing,
Now try x=-inf
f'(x)=(-x^2+1)/(x^2+1)^2)
divide numerator and denominator by 1/x^4
f'(-inf)=-1/1 negative so f(x) is decreasing from -inf to -1
and you can check on the right of 1.

To find the intervals on which the function f(x) = (x)/(x^2+1) is increasing or decreasing, we need to analyze the sign of the derivative, f'(x).

You have correctly calculated the derivative of f(x) as:
f'(x) = (x^2 + 1) - (x)(2x) / (x^2 + 1)^2.

To proceed, we need to simplify the derivative expression:
f'(x) = (x^2 + 1) - 2x^2 / (x^2 + 1)^2
= (x^2 + 1 - 2x^2) / (x^2 + 1)^2
= (1 - x^2) / (x^2 + 1)^2.

Now, we can determine the intervals where f(x) is increasing or decreasing by analyzing the sign of f'(x) in different intervals on the x-axis.

1. Find the critical points by setting f'(x) = 0:
(1 - x^2) / (x^2 + 1)^2 = 0.

For this equation, the numerator can equal zero when:
1 - x^2 = 0, which gives x = Β±1.

However, since (x^2 + 1)^2 is always positive, there are no values for x that make the denominator zero.

2. Determine the intervals:
To analyze the sign of f'(x), we need to choose a test point within each interval. Let's consider x = 0, x = -2, and x = 2.

For x < -1, let's test x = -2:
f'(-2) = (1 - (-2)^2) / ((-2)^2 + 1)^2
= (1 - 4) / (4 + 1)^2
= -3 / 5^2
= -3 / 25.

Since -3 / 25 < 0, f'(x) < 0 for x < -1.

For -1 < x < 1, let's test x = 0:
f'(0) = (1 - 0^2) / (0^2 + 1)^2
= 1 / 1^2
= 1 / 1
= 1.

Since 1 > 0, f'(x) > 0 for -1 < x < 1.

For x > 1, let's test x = 2:
f'(2) = (1 - 2^2) / (2^2 + 1)^2
= (1 - 4) / (4 + 1)^2
= -3 / 5^2
= -3 / 25.

Since -3 / 25 < 0, f'(x) < 0 for x > 1.

3. Summarize the intervals:
Based on the analysis of f'(x), we can conclude:

- f(x) is increasing for -1 < x < 1.
- f(x) is decreasing for x < -1 and x > 1.

Therefore, the intervals on the x-axis where the function f(x) is increasing are -1 < x < 1, and the intervals where it is decreasing are x < -1 and x > 1.