Q. 73 - Calculus

Question

Sketch careful, labeled graphs of each function $f$ in Exercises $57-82$ by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f$ and $f^{\text{'}}$and examine any relevant limits so that you can describe all key points and behaviors of $f$.

$f\left(x\right)=\frac{x^2\left(x-1\right)}{{\left(x-2\right)}^2}$.

Step-by-Step Solution

Answer

The graph for the function $f\left(x\right)=\frac{x^2\left(x-1\right)}{{\left(x-2\right)}^2}$ is,

Step 1 . Given information

$f\left(x\right)=\frac{x^2\left(x-1\right)}{{\left(x-2\right)}^2}$.

Step 2 . Let y = x 2 x - 1 x - 2 2 .

Now point table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$-3$	$\frac{36}{25}$	$\left(-3,\frac{36}{25}\right)$
$-2$	$\frac{3}{4}$	$\left(-2,\frac{3}{4}\right)$
$0$	$0$	$\left(0,0\right)$
$1$	$0$	$\left(1,0\right)$
$3$	$18$	$\left(3,18\right)$

Step 3 . The graph for the function is,

Step 4 . Now for critical point, f ' x = 0 .

$$\begin{gathered} \frac{d}{dx}\left(\frac{x^2\left(x-1\right)}{{\left(x-2\right)}^2}\right)=0 \\ \frac{d}{dx}\left(\frac{\left(x^3-x^2\right)}{{\left(x-2\right)}^2}\right)=0 \\ \frac{{\left(x-2\right)}^2\left(3x^2-2x\right)-\left(x^3-x^2\right)2\left(x-2\right)}{{\left({\left(x-2\right)}^2\right)}^2}=0 \\ \left(x-2\right)\left[\left(x-2\right)\left(3x^2-2x\right)-\left(2x^3-2x^2\right)\right]=0 \\ (x-2)\left[3x^3-2x^2-6x^2+4x-2x^3+2x^2\right]=0 \\ (x-2)\left(x^3-6x^2+4x\right)=0 \\ x(x-2)\left(x^2-6x+4\right)=0 \\ x=0;x=2;x=\frac{-(-6)\pm \sqrt{(-6{)}^2-4·1·4}}{2·1} \\ x=0,2,\frac{6\pm \sqrt{36-16}}{2} \\ x=0,2,\frac{6\pm \sqrt{20}}{2} \\ x=0,2,3\pm \sqrt{5} \end{gathered}$$

Therefore, $f$ has three critical points. It has no local extrema.

Step 5 . The sign chart of f is shown below:

For roots of the function,

$$\begin{gathered} \frac{x^2\left(x-1\right)}{{\left(x-2\right)}^2}=0 \\ {x}^2\left(x-1\right)=0 \\ x=0,1 \end{gathered}$$

Therefore, the function $f$ is defined everywhere except at $x=2$ where there is a vertical asymptote. The function is positive on $(1,2)\cup (2,\infty )$ and negative elsewhere.

The function has local maximum at $x=0$ and local minimum at $x=3\pm \sqrt{5}$.The function is increasing on $(-\infty ,0)\cup (3-\sqrt{5},2)\cup (3+\sqrt{5},\infty )$ and decreasing elsewhere. No horizontal asymptote.