`y = 1/(x(x - 3)^2)` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

Textbook Question

Chapter 4, Review - Problem 23 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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`y=1/(x(x-3)^2)`

a) Asymptotes

Vertical asymptotes are the zeros of the denominator.

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

So, x=0 and x=3 are the vertical asymptotes.

Degree of numerator=0

Degree of denominator=3

Degree of denominator `>` Degree of numerator 

the horizontal asymptote is the x-axis,

So, y=0 is the horizontal asymptote.

b) Maxima/Minima

`f(x)=x^-1(x-3)^-2`

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

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

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

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

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

Now to find critical numbers set f'(x)=0,

`(-3(x-1))/(x^2(x-3)^3)=0 rArrx=1`

Now to find maxima/minima, let's find the sign of f'(x) by plugging in test points in the intervals (-`oo` ,0), (0,1) and (`oo` ,0),

`f'(-1)=(-3(-1-1))/((-1)^2(-1-3)^3)=-3/32`

`f'(0.5)=(-3(0.5-1))/((0.5)^2(0.5-3)^3)=-0.384`

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

So the sign of f'(x) in the interval (0,1) is negative and changes to positive in the interval (`oo` ,0)

Hence, there is Local minima f(1)=1/4 is at x=1

c) Inflection Points

Let's find f''(x) by using quotient rule

`f''(x)=-3(x^2(x-3)^3-(x-1)(x^2(3)(x-3)^2+(x-3)^3(2x)))/(x^4(x-3)^6)`

`f''(x)=((-3(x-3)^2)(x^2(x-3)-(x-1)(3x^2+2x^2-6x)))/(x^4(x-3)^6)`

`f''(x)=(-3(x^3-3x^2-5x^3+6x^2+5x^2-6x))/(x^4(x-3)^4)`

`f''(x)=(-3(-4x^3+8x^2-6x))/(x^4(x-3)^4)`

`f''(x)=(6x(2x^2-4x+3))/(x^4(x-3)^4)`

`f''(x)=(6(2x^2-4x+3))/(x^3(x-3)^4)`

Now let's find inflection points by solving x for f''(x)=0,

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

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

`x=(4+-sqrt(-8))/4`

Since the roots are not real , so there are no inflection points.

 

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