Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 12

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) .