# `y = 1/x^2 - 1/(x - 2)^2` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

### Textbook Question

Chapter 4, Review - Problem 24 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

`y=1/x^2-1/(x-2)^2`

To determine the asymptotes, express the function as a single fraction. Since the LCD of the two fractions is x^2(x-2)^2, then, the function becomes:

`y = 1/x^2 * (x-2)^2/(x-2)^2 - 1/(x-2)^2*x^2/x^2 = (x-2)^2/(x^2(x-2)^2)-x^2/(x^2(x-2)^2)=(x^2 - 4x+4)/(x^2(x-2)^2) - x^2/(x^2(x-2)^2)`

`y=(-4x+ 4)/(x^2(x-2)^2)`

Then, refer to the degree of the numerator and denominator to get the horizontal asymptotes. The degree of the numerator is 1 and the degree of the denominator is 4.  Since the degree of the numerator is less than that of the denominator, thus, the horizontal asymptote is `y=0` .

To get the vertical asymptotes,  take note that in fraction zero denominator is not allowed. So, solve for the values of x that would result to zero denominator.

`x^2(x-2)^2=0`

Set each factor equal to zero.

`x= 0`

`x-2=0`

`x=2`

Hence, the vertical asymptotes are `x=0` and `x=2` .

Next, plot the asymptotes.

Notice that it divides the xy plane into six regions. To determine which region will the curves belong, determine at least two points for the region at the left of x=0, between x=0 and x=2 and at the right of x=2.

For the region at the left of x=0, the points are:

`y=1/(-3)^2 - 1/(-3-2)^2=16/225`     `(-3,16/225)`

`y=1/(-1)^2 - 1/(-1-2)^2=8/9 `        ` (-1,8/9)`

So at the left of x=0, the curve is located above the horizontal asymptote.

For the region between x=0 and x=2, the points are:

`y=1/(1/2)^2-1/(1/2-2)^2=32/9 `         `(1/2,32/9)`

`y=1/1^2-1/(1-2)^2=0 `                       `(1,0)`

At region between the x=0 and x=2, the curve crosses the horizontal asymptote.

And for the region at the right of x=2, the points are:

`y=1/3^2-1/(3-2)^2=-8/9 `          (3,-8/9)

`y=1/5^2-1/(5-2)^2=-16/225 `     `(5,-16/225)`

So at the right of x=2, the curve is located below the horizontal asymptote.

To determine the maximum and minimum points, take the derivative of the function.

`y=1/x^2-1/(x-2)^2 = x^(-2) - (x-2)^(-2)`

`y'=-2x^(-3)-(-2)(x-2)^(-3) `

`y'=-2/x^3 + 2/(x-2)^3`

Then, solve for the critical numbers. To do so, set the derivative equal to zero.

`0=-2/x^3 + 2/(x-2)^3`

`0= -2/x^3*(x-2)^3/(x-2)^3+2/(x-2)^3*x^3/x^3`

`0=(-2(x^3-6x^2+12x-8))/(x^3(x-2)^3)+(2x^3)/(x^3(x-2)^3)`

`0=(-2x^3+12x^2-24x+16)/(x^3(x-2)^3)+(2x^3)/(x^3(x-2)^3)`

`0=(12x^2-24x+16)/(x^3(x-2)^3)`

`0=12x^2-24x+16`

`0=3x^2-6x+4`

Apply quadratic formula to solve for x.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-6)+-sqrt((-6)^2-4(3)(4)))/(2*3)=(6+-sqrt(-12))/6`

Since the number inside the square root is negative, the values of x are non-real number. It means that there are no values of x that will result to y'=0.

However, to get the critical numbers,  consider also the values of x which make the first derivative undefined. y' is undefined when its denominator is equal to zero.

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

Set each factor equal to zero.

` `

`x=0`

`(x-2)^3=0`

`x-2=0`

`x=2`

So the critical numbers are x=0 and x=2. Notice that these are also the vertical asymptotes. This means that the function has no maximum or minimum points.

Next, determine the intervals in which the function is increasing and decreasing.

Take note that the boundaries of the interval in which the function is increasing or decreasing are the critical numbers x=0 and x=2. So the intervals that should be considered are `(-oo, 0)` , `(0,2)` and `(2,oo)` .

Then, take the a test value for each interval. Plug-in them to the first derivative.

`y'=-2/x^3 + 2/(x-2)^3`

If the result of y' is negative, the function is decreasing in that interval. If it is positive,  the function is increasing in that interval.

For the first interval `(-oo, 0)` , let the test value be x= -1.

`y'=-2/(-1)^3+2/(-1-2)^3=2-2/27=52/27`  (increasing)

For the second interval (0,2), let the test value of be x=1.

`y'=-2/1^3+2/(1-2)^3=-4 `  (decreasing)

And for the third interval `(2,oo)` , let the test value  be x=3.

`y'=-2/3^3+2/(3-2)^3=52/27`   (increasing)

Thus, the function is increasing at intervals `(-oo, 0) uu (2,oo)` . And it is decreasing at interval `(0,2)` .

Next, determine the inflection. So take the second derivative of the function.

`y'=-2/x^3 + 2/(x-2)^3=-2x^(-3)+2(x-2)^(-3)`

`y''=6x^(-4)-6(x-2)^(-4)`

`y''=6/x^4-6/(x-2)^4`

Then, set the second derivative equal to zero.

`0=6/x^4-6/(x-2)^4`

`0=6/x^4*(x-2)^4/(x-2)^4 - 6/(x-2)^4*x^4/x^4`

`0=(6(x-2)^4)/(x^4(x-2)^4)- (6x^4)/(x^4(x-2)^4)`

`0=(-48x^3+144x^2-192x+96)/(x^4(x-2)^4)`

`0=-48x^3+144x^2-192x+96`

`0=-48(x^3-3x^2+4x-2)`

`0=x^3-3x^2+4x-2`

To solve, factor it by grouping.

`0=x^3-1 - 3x^2+4x - 1`

`0=(x^3-1) - (3x^2-4x+1)`

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

`0=(x-1)[(x^2+x+1)-(3x-1)]`

`0=(x-1)(x^2-2x+2)`

Here,  consider only the factor (x - 1).  It is because setting the factor (x^2-2x+2) equal to zero would result to a non-real number.

`0=x - 1`

`1=x`

Hence, the change of concavity occurs only at  x=1.

Next, determine the concavity of the intervals `(-oo, 0)` , `(0,1)` , `(1,2) ` and `(2,oo)` . To do so, assign a test value for each interval and plug-in them to the second derivative.

`y''=6/x^4-6/(x-2)^4`

Take note that if the resulting value of y" is positive, the function is concave up in that interval. And if the resulting value of y" is negative, the function is concave down.

For the first interval `(-oo,0)` , let the test be x=-1.

`y''=6/(-1)^4-6/(-1-2)^4=160/27` (concave up)

For the second interval `(0,1)` , let the test value be x=1/2.

`y''=6/(1/2)^2-6/(1/2-2)^4=2560/27`  (concave up)

For the third interval (1,2), let the test value be x=3/2.

`y''=6/(3/2)^4-6/(3/2-2)^4=-2560/27 `   (concave down)

And for the last interval `(2,oo)` , let the test value be x=3.

`y''=6/3^4-6/(3-2)^2=-160/27`   (concave down).

Thus, the function is concave upward in the intervals `(-oo,0) uu (0,1)` . And it is concave downward in the intervals `(1,2) uu (2,oo)` .

Therefore, the graph of the function `y = 1/x^2-1/(x-2)^2` is: