Chapter 4.3, Problem 38E

(a)

To determine

To find : the point at which $y=f\left(x\right)$has a local maximum.

Answer to Problem 38E

One local maximum occur at $x=2$ .

Explanation of Solution

Given information :

  $y^{\prime }={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$

Formula needed :

First derivative test

The domain of $y^{\prime }$ is $\left(-\infty ,\infty \right)$ , so the domain of $y$ is also $\left(-\infty ,\infty \right)$ .

The critical points of $y$ occur only at the zeros of $y^{\prime }$ .

Since, $y^{\prime }={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$ .

The first derivative is zero at $x=1,x=2$ and, $x=4$ .

The behavior table of $y^{\prime }$ is

  $\begin{array}{ccccc}\text{Interval}& x<1& 1<x<2& 2<x<4& 4<x\\ \text{sign}\text{of}{y}^{\prime }& +& +& -& +\\ \text{Behavior}\text{of}y& \text{Increasing}& \text{Increasing}& \text{Decreasing}& \text{Increasing}\end{array}$

Using the first derivative test and the above table, there is one local maximum occur at $x=2$ .

(b)

To determine

To find : the point at which $y=f\left(x\right)$has a local minimum.

Answer to Problem 38E

One local minimum occur at $x=4$ .

Explanation of Solution

Given information :

  $y^{\prime }={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$

Formula needed :

First derivative test

The domain of $y^{\prime }$ is $\left(-\infty ,\infty \right)$ , so the domain of $y$ is also $\left(-\infty ,\infty \right)$ .

The critical points of $y$ occur only at the zeros of $y^{\prime }$ .

Since, $y^{\prime }={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$ .

The first derivative is zero at $x=1,x=2$ and, $x=4$ .

The behavior table of $y^{\prime }$ is

  $\begin{array}{ccccc}\text{Interval}& x<1& 1<x<2& 2<x<4& 4<x\\ \text{sign}\text{of}{y}^{\prime }& +& +& -& +\\ \text{Behavior}\text{of}y& \text{Increasing}& \text{Increasing}& \text{Decreasing}& \text{Increasing}\end{array}$

Using the first derivative test and the above table, there is one local minimum occur at $x=4$ .

(c)

To determine

To find : the point at which $y=f\left(x\right)$has a point of inflection.

Answer to Problem 38E

The inflection points of $y=f\left(x\right)$ at $x=1,x=1.634$ and, $x=3.366$ .

Explanation of Solution

Given information :

  $y^{\prime }={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$

Formula needed :

second derivative test

The domain of $y^{\prime }$ is $\left(-\infty ,\infty \right)$ , so the domain of $y$ is also $\left(-\infty ,\infty \right)$ .

Differentiate $y^{\prime }={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$ with respect to $x$ ,

  $\begin{array}{c}{y^{\prime }}^{\prime }=\frac{d}{dx}{\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)\\ =\left(x-2\right)\left(x-4\right)\cdot \frac{d}{dx}{\left(x-1\right)}^2+{\left(x-1\right)}^2\cdot \frac{d}{dx}\left(x-2\right)\left(x-4\right)\\ =\left(x-2\right)\left(x-4\right)\cdot \frac{d}{dx}{\left(x-1\right)}^2+{\left(x-1\right)}^2\cdot \left[\left(x-4\right)\cdot \frac{d}{dx}\left(x-2\right)+\left(x-2\right)\cdot \frac{d}{dx}\left(x-4\right)\right]\\ =\left(x-2\right)\left(x-4\right)\cdot \frac{d}{dx}{\left(x-1\right)}^2+{\left(x-1\right)}^2\left(x-4\right)\cdot \frac{d}{dx}\left(x-2\right)+{\left(x-1\right)}^2\left(x-2\right)\cdot \frac{d}{dx}\left(x-4\right)\end{array}$

Take factor common,

  $\begin{array}{c}{y^{\prime }}^{\prime }=\left(x-2\right)\left(x-4\right)\cdot 2\left(x-1\right)+{\left(x-1\right)}^2\left(x-4\right)\cdot \left(1\right)+{\left(x-1\right)}^2\left(x-2\right)\cdot \left(1\right)\\ =2\left(x-2\right)\left(x-4\right)\left(x-1\right)+{\left(x-1\right)}^2\left(x-4\right)+{\left(x-1\right)}^2\left(x-2\right)\\ =\left(x-1\right)\left[2\left(x-2\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)+\left(x-1\right)\left(x-2\right)\right]\\ =\left(x-1\right)\left[\left(x-4\right)\left[2\left(x-2\right)+\left(x-1\right)\right]+\left(x-1\right)\left(x-2\right)\right]\end{array}$

Simplify further,

  $\begin{array}{c}{y^{\prime }}^{\prime }=\left(x-1\right)\left[\left(x-4\right)\left(3x-5\right)+\left(x-1\right)\left(x-2\right)\right]\\ =\left(x-1\right)\left[3x^2-5x-12x+20+x^2-3x+2\right]\\ =\left(x-1\right)\left[4x^2-20x+22\right]\\ =\left(x-1\right)\left(x-1.634\right)\left(x-3.366\right)\end{array}$

The inflection points of $y$ occur only at the zeros of ${y^{\prime }}^{\prime }$ .

Since, ${y^{\prime }}^{\prime }=\left(x-1\right)\left(x-1.634\right)\left(x-3.366\right)$ .

The second derivative is zero at $x=1,x=1.634$ and, $x=3.366$ .