Chapter 4, Problem 16RE

a.

To determine

To identify: The interval in which the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is increasing.

Answer to Problem 16RE

The function is increasing in the interval $\left(-\infty ,0.215\right]$ .

Explanation of Solution

Given information:

The given function is $y=\frac{5-4x+4x^2-x^3}{x-2}$

Consider the given function.

Calculate the derivative of the function.

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{\left(x-2\right)\left(-4+8x-3x^2\right)-\left(5-4x+4x^2-x^3\right)}{{\left(x-2\right)}^2}\\ =\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}\end{array}$

Now determine the critical points.

By putting $\frac{dy}{dx}=0$ .

  $\begin{array}{c}\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}=0\\ -2x^3+10x^2-16x+3=0\end{array}$

So, the critical points are: 2 and 0.215.

The required intervals are: $\left(-\infty ,0.215\right),\left(0.215,2\right)\text{ and }\left(2,\infty \right)$

Now draw the table 1:

Intervals	$\left(-\infty ,0.215\right)$	$\left(0.215,2\right)$	$\left(2,\infty \right)$
Sign of derivative	$+$	$-$	$-$
Behavior of y	Increasing	Decreasing	Decreasing

Thus, from the table 1 it can be observed that the function is increasing in the interval $\left(-\infty ,0.215\right]$ .

b.

To determine

To identify: The interval in which the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is decreasing.

Answer to Problem 16RE

The function is decreasing in the interval $\left(0.215,2\right)$ , and $\left(2,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{5-4x+4x^2-x^3}{x-2}$

It is known that the table 1 is:

Intervals	$\left(-\infty ,0.215\right)$	$\left(0.215,2\right)$	$\left(2,\infty \right)$
Sign of derivative	$+$	$-$	$-$
Behavior of y	Increasing	Decreasing	Decreasing

Thus, from the table 1 it can be observed that the function is decreasing in the interval $\left(0.215,2\right)$ , and $\left(2,\infty \right)$ .

c.

To determine

To identify: The interval in which the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is concave up.

Answer to Problem 16RE

The function is concave up in the interval $\left(2,3.710\right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{5-4x+4x^2-x^3}{x-2}$

Consider the given function.

Determine the second derivative and equate with zero.

  $\begin{array}{c}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[\frac{\left(x-2\right)\left(-4+8x-3x^2\right)-\left(5-4x+4x^2-x^3\right)}{{\left(x-2\right)}^2}\right]\\ =\frac{-2\left(x^3-6x^2+12x-13\right)}{{\left(x-2\right)}^3}\\ 0=\frac{-2\left(x^3-6x^2+12x-13\right)}{{\left(x-2\right)}^3}\\ 0=x^3-6x^2+12x-13\end{array}$

So, the points where double derivative is zero $x=3.710$ .

Also, function is not defined at $x=2$ .

Now draw the table 2:

Intervals	$x<2$	$2<x<3.710$	$3.710<x$
Sign of derivative	$-$	$+$	$-$
Behavior of y	Concave down	Concave up	Concave down

From the table 2 it can be observed that the function is concave up in the interval $\left(2,3.710\right)$ .

d.

To determine

To identify: The interval in which the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is concave down.

Answer to Problem 16RE

The function is concave down in the interval $\left(-\infty ,2\right)$ , and $\left(3.710,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{5-4x+4x^2-x^3}{x-2}$

Consider the given function.

It is known that the table 2 is:

Intervals	$x<2$	$2<x<3.710$	$3.710<x$
Sign of derivative	$-$	$+$	$-$
Behavior of y	Concave down	Concave up	Concave down

From the table 2 it can be observed that the function is concave down in the interval $\left(-\infty ,2\right)$ , and $\left(3.710,\infty \right)$ .

Now draw the graph of the function.

  

Thus, answer is verified from the graph.

e.

To determine

To identify: The local extreme values of the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ .

Answer to Problem 16RE

Local maxima at $\left(0.215,-2.417\right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{5-4x+4x^2-x^3}{x-2}$

Consider the given function.

The value of y at these critical points $x=0.215$ is:

  $\frac{5-4\left(0.215\right)+4{\left(0.215\right)}^2-{\left(0.215\right)}^3}{\left(0.215\right)-2}=-2.417$

At $x=2$ the value is:

  $\frac{5-4\left(2\right)+4{\left(2\right)}^2-{\left(2\right)}^3}{\left(2\right)-2}=\text{Not exist}$

Thus, local maxima at $\left(0.215,-2.417\right)$ .

f.

To determine

To identify: The inflection points of the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ .

Answer to Problem 16RE

The inflection point is: $\left(3.710,-3.420\right)$

Explanation of Solution

Given information:

The given function is $y=\frac{5-4x+4x^2-x^3}{x-2}$

Consider the given function.

From the table it can be observed that there is only one inflation point which is: $\left(3.710,-3.420\right)$ .