Steps for Sketching the Graph of the Function

Example 1. Sketch graph of the function `f(x)=1/(x-1)`.

Since `f(x)=g(x-1)` where `g(x)=1/x` then we study function `g(x)=1/x` and then shift graph one unit to the right.

Function is odd because `g(x)=-g(-x)`, so we consider only interval `[0,oo)` and then just reflect result about the origin. Function is not periodic.

There is no y-intercept because `g(0)` is undefined.

There is no x-intercept because there are no points where `g(x)=1/x=0`.

Since `lim_(x->0^+)1/x=+oo` then `x=0` is vertical asymptote.

Since `lim_(x->oo)1/x=0` then `y=0` is horizontal asymptote.

We have that `g'(x)=-1/x^2`. There are no points where derivative equals 0. Point `x=0` is where derivative doesn't exist.

Since `g'(x)=-1/x^2<0` for all `x` then function is decreasing on `(0,+oo)` and there are no extrema according to First Derivative Test.

Second derivative is `g''(x)=2/x^3`. There are no points where `g''(x)=0` and second derivative doesn't exist when `x=0`.

Since second derivative is positive for `x>0` then function is concave upward on `(0,oo)`.

Point `x=0` is not point of inflection because `x=0` is not in the domain of the function.

So, we actually don't have important points, but when `x` approaches 0 from the right function grows without a bound and we can sketch function, taking into account that functions is decreasing and concave upward on `(0,+oo)`. For better graph take a couple of control points: `(1,1),(1,1/2),(4,1/4)`.

Since function is odd, reflect it about origin.

Shift the function 1 unit to the right.

Example 2. Sketch graph of the function `f(x)=(x+2)^2(x-1)^3`.

There is no simpler function that initial function is obtained from.

Function is neither even nor odd and not periodic.

`f(0)=(0+2)^2(0-1)^3=-4`, so y-intercept is -4.

x-intercepts are points where `f(x)=0`: `(x+2)^2(x-1)^3=0` which gives `x=-2` and `x=1`.

Since function is continuous everywhere, `lim_(x->-oo)f(x)=-oo` and `lim_(x->oo)f(x)=oo` then there are no asymptotes.

We have that `f'(x)=2(x+2)(x-1)^3+3(x+2)^2(x-1)^2=(x+2)(x-1)^2(2x-2+3x+6)=`

`=(x+2)(x-1)^2(5x+4)`.

So, derivative equals 0 when `x=-2,x=1,x=-4/5`.

According to method of intervals number line is divided by stationary points on four intervals.

First interval is `(-oo,-2)`: taking any point from this interval, for example `x=-3` we obtain that `f'(-3)=(-3+2)(-3-1)^2(5*(-3)+4)>0`, so `f'(x)>0` on this interval.

Second interval is `(-2,-4/5)`: taking any point from this interval, for example `x=-1` we obtain that `f'(-1)=(-1+2)(-1-1)^2(5*(-1)+4)<0`, so `f'(x)<0` on this interval.

Third interval is `(-4/5,1)`: taking any point from this interval, for example `x=0` we obtain that `f'(0)=(0+2)(0-1)^2(5*(0)+4)>0`, so `f'(x)>0` on this interval.

Fourth interval is `(1,oo)`: taking any point from this interval, for example `x=3` we obtain that `f'(3)=(3+2)(3-1)^2(5*3+4)>0`, so `f'(x)>0` on this interval.

According to First Derivative Test `x=-2` is local maximum, `x=-4/5` is local minimum and `x=1` is not an extremum.

Second derivative is `f''(x)=2(x-1)(10x^2+16x+1)`.

Roots of the equation `10x^2+16x+1=0` are `x=(-8+3sqrt(6))/10~~-0.065` and `x=(-8-3sqrt(6))/10~~-1.535`.

So, points where second derivative equals 0 are `x=1,x=-0.065,x=-1.535`.

Clearly they are all points of inflections because second derivative changes sign at these points.

`f''(x)>0` on `(-1.535,-0.065)uu(1,oo)` and `f''(x)<0` on `(-oo,-1.535)uu(-0.0065,1)`.

So we have following "important" points:

Function is increasing on `(-oo,-2)uu(-4/5,oo)` and decreasing on `(-2,-4/5)`.

Function is concave upward on `(-1.535,-0.065)uu(1,oo)` and concave downward on `(-oo,-1.535)uu(-0.0065,1)`.

Note that points `(-0.065,-4.052)` and `(0,-4)` are very close, so they are almost indistinguishable on graph.

Example 3. Sketch graph of the function `f(x)=(x^2-5x+6)/(x^2+1)`.

There is no simpler function that initial function is obtained from.

Function is neither even nor odd and not periodic.

`f(0)=(0^2-5*0+6)/(0^2+1)=6`, so y-intercept is 6.

x-intercepts are points where `f(x)=0`: `(x^2-5x+6)=(x-2)(x-3)=0` which gives `x=2` and `x=3`.

Since function is continuous everywhere, `lim_(x->oo)(x^2-5x+6)/(x^2+1)=1` then there is only one asymptote: horizontal asymptote `y=1`.

We have that `f'(x)=((x^2-5x+6)'(x^2+1)-(x^2-5x+6)(x^2+1)')/(x^2+1)^2=5(x^2-2x-1)/(x^2+1)^2`.

Roots of the equation `x^2-2x-1=0` are `x=1+sqrt(2)~~2.41` and `x=1-sqrt(2)~~-0.41`.

So, derivative equals 0 when `x=-0.41,x=2.41`.

According to method of intervals number line is divided by stationary points on three intervals.

First interval is `(-oo,-0.41)`: `f'(x)>0` on this interval.

Second interval is `(-0.41,2.41)`:`f'(x)<0` on this interval.

Third interval is `(2.41,oo)`: `f'(x)>0` on this interval.

According to First Derivative Test `x=-0.41` is local maximum, `x=2.41` is local minimum.

Second derivative is `f''(x)=-10((x+1)(x^2-4x+1))/(x^2+1)^3`.

Roots of the equation `x^2-4x+1=0` are `x=2-sqrt(3)~~0.27` and `x=2+sqrt(3)~~3.73`.

So, points where second derivative equals 0 are `x=-1,x=0.27,x=3.73`.

Clearly they are all points of inflections because second derivative changes sign at these points.

`f''(x)>0` on `(-oo,-1)uu(0.27,3.73)` and `f''(x)<0` on `(-1,0.27)uu(3.73,oo)`.

So we have following "important" points:

Function is increasing on `(-oo,-0.41)uu(2.41,oo)` and decreasing on `(-0.41,2.41)`.

Function is concave upward on `(-oo,-1)uu(0.27,3.73)` and concave downward on `(-1,0.27)uu(3.73,oo)`.

Horizontal asymptote is `y=1`.

Example 4. Sketch graph of the function `f(x)=(x-1)^3/(x+1)^2`.

There is no simpler function that initial function is obtained from.

Function is neither even nor odd and not periodic.

`f(0)=(0-1)^3/(0+1)^2=-1`, so y-intercept is -1.

x-intercepts are points where `f(x)=0`: `(x-1)^3=0` which gives `x=1`.

Vertical asymptote is `x=-1` because `lim_(x->-1^+)(x-1)^3/(x+1)^2=-oo`.

There are no horizontal asymptotes because `lim_(x->oo)(x-1)^3/(x+1)^2=oo` and `lim_(x->-oo)(x-1)^3/(x+1)^2=-oo`.

Since `lim_(x->oo)(f(x))/x=lim_(x->oo)(x^3-3x^2+3x-1)/(x^3+2x^2+x)=1` then `m=1`. Now `lim_(x->oo)(f(x)-mx)=(x^3-3x^2+3x-1)/(x^2+2x+1)-x=lim_(x->oo)(-5x^2+2x-1)/(x^2-2x+1)=-5`.

So, there is slant asymptote `y=x-5`.

We have that `f'(x)=(((x-1)^3)'(x+1)^2-(x-1)^3((x+1)^2)')/(x+1)^4=((x-1)^2(x+5))/(x+1)^3`.

Derivative equals 0 when `x=1,x=-5` and doesn't exist when `x=-1`.

Since `(x-1)^2>=0` then these factor doesn't influence sign of derivative.

Thus `f'(x)>0` when `x in (-oo,-5)uu(-1,oo)` and `f'(x)<0` when `x in (-5,-1)`.

According to First Derivative Test `x=-5` is local maximum and `x=1` is not an extremum.

Point `x=-1` is not in the domain of function, so clearly not an extremum.

Second derivative is `f''(x)=24 (x-1)/(x+1)^4`.

So, point where second derivative equals 0 is `x=1`.

Clearly it is inflection point because second derivative changes sign at this point.

`f''(x)>0` when `x>1` and `f''(x)<0` when `x<1`.

Let's add a couple of control points:

`x=3`, `f(3)=0.5`.

`x=-8`, `f(-8)~~-14.88`.

So, we have following points:

Function is increasing on `(-oo,-5)uu(-1,oo)` and decreasing on `(-5,-1)`.

Function is concave upward on `(1,oo)` and concave downward on `(-oo,1)`.

Vertical asymptote is `x=-1` and slant asymptote is `y=x-5`.