Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 23

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.