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.
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...
2 answers
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.
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.
1 answer