curve sketching: f(x) = x / (x - 1)^2

G

Guest

Guest
f(x) = x / (x -1 )^2

DOMAIN
x cannot equal 1

INTERCEPTS
x-int
0=x/(x-1)^2
x=0

y-int
y=0/(0-1)^2
y=0

SYMMETRY
f(-x)=-x/(-x-1)^2
-no symmetry

-f(-x)=-(x/(x-1)^2
=-

ASYMPTOTES


...then im supposed to find the max/min values and the concavity which im pretty sure i did correctly however im not sure if i did the stuff up there correctly. Thanks for any help.
 
Hello, bandaid-bandet!

f(x)=x(x1)2\displaystyle f(x) \:= \:\frac{x}{(x -1 )^2}

DOMAIN: x1\displaystyle \:x\,\neq\,1

INTERCEPTS
x-int: 0=x(x1)2        x=0\displaystyle \:0\:=\:\frac{x}{(x-1)^2}\;\;\Rightarrow\;\;x\,=\,0

y-int: y=0(01)2    y=0\displaystyle \:y\:=\:\frac{0}{(0-1)^2}\;\;y\,=\,0

SYMMETRY
f(x)=x(x1)2\displaystyle f(-x)\:=\:\frac{-x}{(-x-1)^2} . . . no symmetry

f(x)=x(x1)2\displaystyle \,-f(-x)\:=\:-\frac{x}{(x-1)^2} . . . no symmetry

Then I'm supposed to find the max/min values and the concavity
. . which I'm pretty sure I did correctly.
However, I'm not sure if I did the stuff up there correctly. . You did great!

ASYMPTOTES

You already found the vertical asymptote: x=1\displaystyle \,x\,=\,1

Horizontal asymptote: \(\displaystyle \L\:\lim_{x\to\infty}y\;=\;\lim_{x\to\infty}\frac{x}{(x\,-\,1)^2}\:=\:0\)
. . Hence: y=0\displaystyle \:y\,=\,0 (x-axis)


MAX\MIN
We have: f(x)=x+1(x1)3        x=1\displaystyle \:f'(x)\:=\:-\frac{x\,+\,1}{(x\,-\,1)^3}\;\;\Rightarrow\;\;x\,=\,-1 . . . critical value

And we have: f(x)=2(x+2)(x1)4\displaystyle \:f''(x)\:=\:\frac{2(x\,+\,2)}{(x\,-\,1)^4}

At x=1:    f(1)=2(1+2)(11)4=+18\displaystyle x\,=\,-1:\;\;f''(-1)\:=\:\frac{2(-1\,+\,2)}{(-1\,-\,1)^4}\:=\:+\frac{1}{8} . . . positive, concave up: \displaystyle \,\cup

. . Hence, a minimum at (1,14)\displaystyle \left(-1,\,-\frac{1}{4}\right)


CONCAVITY
We have: f(x)=2(x+2)(x1)4\displaystyle \:f''(x)\:=\:\frac{2(x\,+\,2)}{(x\,-\,1)^4}

Inflection point at x=2,  y=29\displaystyle x\,=\,-2,\;y\,=\,-\frac{2}{9}

When x<2:    f(x)<0\displaystyle x\,<\,-2:\;\;f''(x) \,<\,0 . . . concave down
When x>2:    f(x)>0\displaystyle x\,>\,2:\;\;f''(x)\,>\,0 . . . concave up


The graph looks like this:
Code:
                                |           :
                                |          *:*
                                |           :
                                |         * : *
                                |        *  :  *
                                |      *    :     *
           -2   -1              |   *       :           *
    --------+----+--------------*-----------+-------------- 
     *                      *   |           :1
            *          *        |           :
                 *              |           :
                                |           :
 
Top