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

Expert Answers
gsarora17 eNotes educator| Certified Educator


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







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),




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







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




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


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