# Find min/max along with concavity, inflection points, and asymptotes

For the function $ƒ (x)=\frac{x^2}{(x−2)^2}$

I know the derivative is $f'(x)=-\frac{4x}{\left(x-2\right)^3}$ and $f''(x) =\frac{8\left(x+1\right)}{\left(x-2\right)^4}$

Critical point is $x=-1$ which is also the min.

I think possible inflection point are $x=1$ and $x=0$

I need to find: 1)Find the vertical and horizontal asymptotes of $ƒ (x)$.

2)Find the intervals on which $ƒ (x)‍$ is increasing or decreasing.

3)Find the critical numbers and specify where local maxima and minima of $ƒ (x)$ occur.

4)Find the intervals of concavity and the inflection points of $ƒ (x)$.

sorry for the long problem. Hope you can help

• The critical point should be $x = 0$ (when $f'(x) = 0$), at which $f$ takes on its minimum value. – Namaste Nov 8 '14 at 20:04

we have $f''(x)=8\,{\frac {x+1}{ \left( x-2 \right) ^{4}}}$ and $f''(x)=0$ if $x=-1$ and $f'''(x)=-24\,{\frac {x+2}{ \left( x-2 \right) ^{5}}}$ plugging $x=-1$ in $f'''(x)$ we get $f'''(-1)=\frac{8}{81}\ne 0$ thus we have the inflection point $-1,\frac{1}{9}$ $x=0$ gives no inflection point. We get the point $(0,0)$ as a local minimum. Since $f(x)=1+\frac{4}{x-2}+\frac{4}{(x-2)^2}$ we get a horizontal asymptote $y=1$. Since for $x=2$ is the denominator equal to zero we have a vertical asymptote for $x=2$ we get for $-\infty<x<0$ and $2<x<\infty$ is the given function monotonnously decreasing and for $0<x<2$ monotonously increasing.
• oh wow thanks I think I totally did this wrong then. So let me just make sure I understand. The local min is $x=0$ and the inflection points are $-1 and \frac{1}{9}$ Was I right that there is no local max? – CSCI_newbie Nov 8 '14 at 20:18
• $x=-1$ is the x-coordinate and $y=1/9$ is the y-coordinate of the inflection point – Dr. Sonnhard Graubner Nov 8 '14 at 20:21