`f(x) = x/(x^2+1)`
(a) Take the derivative of the given function.
`f'(x) = ((x^2+1)(1) - (x)(2x))/(x^2+1)^2= (x^2+1-2x^2)/(x^2+1)^2`
`f'(x)=(1-x^2)/(x^2+1)^2`
Then, solve for the critical numbers by setting the derivative equal to zero.
`0=(1-x^2)/(x^2+1)^2`
`0=1 - x^2`
`0=(1 - x)(1 + x)`
`x=-1`
`x=1`
So the critical numbers are x=-1 and x=1....
See
This Answer NowStart your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
`f(x) = x/(x^2+1)`
(a) Take the derivative of the given function.
`f'(x) = ((x^2+1)(1) - (x)(2x))/(x^2+1)^2= (x^2+1-2x^2)/(x^2+1)^2`
`f'(x)=(1-x^2)/(x^2+1)^2`
Then, solve for the critical numbers by setting the derivative equal to zero.
`0=(1-x^2)/(x^2+1)^2`
`0=1 - x^2`
`0=(1 - x)(1 + x)`
`x=-1`
`x=1`
So the critical numbers are x=-1 and x=1. The intervals formed by these two critical numbers are
`(-oo,-1)` ` (-1,1)` and `(1,oo)` .
Then, assign a test value for each interval and plug-in them to the first derivative.
`f'(x)=(1-x^2)/(x^2+1)^2`
If the resulting value of f'(x) is negative, the function is decreasing in that interval. If it is positive, the function is increasing.
For our first interval` (-oo,-1)` , let the test value be x=-2.
`f'(-2) = (1-(-2)^2)/((-2)^2+1)^2=-3/25` (Decreasing)
For our second interval (-1,1), let the test value be x=0.
`f'(0)=(1-0^2)/(0^2+1)^2=1` (Increasing)
And for our third interval, let the test value be x=2.
`f(2)= (1-2^2)/(2^2+1)^2=-3/25` (Decreasing)
Therefore, the function is decreasing at `(-oo, -1) uu (1, oo)` . And it is increasing at the interval `(-1,1)` .
(b) Let's determine at what values of x do the local maximum and minimum occur. To do so, refer to the change of signs of f'(x) before and after the critical number.
Before and after the critical number x=-1, the f'(x) changes from negative values to positive values. So the function has a local minimum at x=-1. The local minimum value of f(x) is:
`f(-1) = (-1)/((-1)^2+1)=-1/2`
Also, before and after the critical number x=1, the f'(x) changes from positive to negative. So, the function has a local maximum at x=1. The local maximum value of f(x) is:
`f(1)=1/(1^2+1)=1/2`
Therefore, the local minimum is `f(x) = -1/2` , which occurs at x=-1. And its local maximum is` f(x)=1/2` , which occurs at x=1.
(c) To determine the concavity of the function, take the second derivative of the function.
`f'(x) = (1 -x^2)/(x^2+1)^2`
`f''(x) = ((x^2+1)^2(-2x) - (1-x^2)(2)(x^2+1)(2x))/(x^2+1)^4`
`f''(x) = (2x(x^2+1)(-(x^2+1)-2(1-x^2)))/(x^2+1)^4`
`f''(x) = (2x(x^2+1)(-x^2-1-2+2x^2))/(x^2+1)^4`
`f''(x)=(2x(x^2+1)(x^2-3))/(x^2+1)^4`
`f''(x)= (2x(x^2-3))/(x^2+1)^3`
Then, set the second derivative equal to zero.
`0=2x(x^2-3)`
Set each factor equal to zero and isolate the x.
For the first factor:
`2x = 0`
`x=0`
For the second factor:
`x^2 - 3=0`
`x^2=3`
`x=+-sqrt3`
Thus, the function changes concavity at `x=-sqrt3` , `x=0` and `x=sqrt3` . So the different concavity occurs are
`(-oo, -sqrt3)` `(-sqrt3,0) ` `(0,sqrt3)` and `(sqrt3, oo)` .
To determine at which interval is the function concave upward or downward, assign a test value. Plug-in them to the second derivative.
`f''(x)= (2x(x^2-3))/(x^2+1)^3`
If the resulting value of f''(x) is negative, the function is concave downward in that interval. If it is positive, the function is concave upward.
For the first interval `(-oo, -sqrt3)` , let the test value be x=-2.
`f''(-2)=(2(-2)[(-2)^2-3])/[(-2)^2+1]^3 =-4/125` (Downward)
For the second interval `(-sqrt3, 0)` , let the test value be x=-1.
`f''(-1)=(2(-1)[(-1)^2-3])/[(-1)^1+1]^3=1/2` (Upward)
For the third interval `(0, sqrt3)` , let the test value be x=1.
`f''(1)=(2(1)(1^2-3))/(1^2+1)^3=-1/2` (Downward)
And for the fourth interval `(sqrt3, oo)` , let the test value be x=2.
`f''(2)= (2(2)(2^2-3))/(2^2+1)^3=4/125` (Upward)
Therefore, the function is concave downward at `(-oo,-sqrt3) uu (0,sqrt3` ). And it is concave upward at `(-sqrt3,0) uu (sqrt3,oo)` .
To get the inflection points, plug-in `x=-sqrt3` , `x=0` and `x=sqrt3` to the original function.
`f(x)=x/(x^2+1)`
`f(-sqrt3)=(-sqrt3)/((-sqrt3)^2+1)=-sqrt3/4`
`f(0)=0/(0^2+1)=0`
`f(sqrt3)=sqrt3/((sqrt3)^2+1)=sqrt3/4`
Hence, the inflection points are `(-sqrt3, -sqrt3/4)` , `(0,0)` and `(sqrt3,sqrt3/4)` .