`f(x) = x/(x^2 + 1)` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the...

`f(x) = x/(x^2 + 1)` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the inflection points.

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Chapter 4, 4.3 - Problem 12 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lemjay | High School Teacher | (Level 3) Senior Educator

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`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)` .

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