`f(x) = 4/(x^2 + 1)` Find the points of inflection and discuss the concavity of the graph of the function.

Textbook Question

Chapter 3, 3.4 - Problem 23 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

2 Answers | Add Yours

nees101's profile pic

nees101 | Student, Graduate | (Level 2) Adjunct Educator

Posted on

Given the function `f(x)=4/(x^2+1)`

Taking the first derivative we get,

`f'(x)=(-8x)/(x^2+1)^2`

Again differentiating we get,

`f''(x)=(-8(x^2+1)^2+8x(4x^3+4x))/(x^2+1)^4`

          `=(-8(x^2+1)+32x^2)/(x^2+1)^3`

          `=(8(3x^2-1))/(x^2+1)^3`

Now the points of inflection can be found out by equation f''(x) to zero.i.e.

f''(x)=0

i.e. `(8(3x^2-1))/(x^2+1)^3=0`

     `8(3x^2-1)=0`

Implies  ````

So, these are the points of inflection. Now choose auxilliary values x=-1 to the left , x=0 and x=1 to the right of the inflection points. ``

So, f''(-1)=2>0. Therefore on `(-infty,-sqrt(1/3))`

the curve is concave upward.

Now f''(0)=-8<0. Therefore on `(-sqrt(1/3),sqrt(1/3))`

the curve is concave downward.

Again f''(1)=2>0. Therefore on `(sqrt(1/3),infty)`

the curve is concave upward.

1 reply Hide Replies

loves2learn's profile pic

loves2learn | (Level 3) Salutatorian

Posted on

Find the second derivative.

Therefore, given

`f(x)=4/(x^2+1) `

`f'(x)=(-8x)/(x^2+1)^2 `

and

`f''(x)=(8(3x^2-1))/(x^2+1)^3 `

Now we need to find the value(s) of x that make the second derivative zero:

`0=8(3x^2-1) `

`0=24x^2-8 `

`8=24x^2 `

`x^2=1/3 `

`x=+-sqrt(1/3) `

Try x values on both sides

So for `x=-sqrt(1/3) ` try `-1 ` and `0 `

`f''(-1)=2 `

and

`f''(0)=-8 `

The concavity does change so `x=-sqrt(1/3) ` is a point of inflection

Find the y-value for this x value, `(-sqrt(1/3),3) `

Try values on both sides for `x=sqrt(1/3) ` as well, like 0 and 1

`f''(0)=-8 `

and

`f''(1)=2 `

The concavity changes around this point as well, therefore it is an inflection point.

Find the y-value for the x value `(sqrt(1/3),3) `

We’ve answered 318,983 questions. We can answer yours, too.

Ask a question