Chapter 4.3, Problem 37E

(a)

To determine

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

Answer to Problem 37E

No local maximum occurs at any critical points.

Explanation of Solution

Given information :

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

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

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

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

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

Using the first derivative test and the above table, there is no local maximum occur at any critical points.

(b)

To determine

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

Answer to Problem 37E

One local minimum occur at $x=2$ .

Explanation of Solution

Given information :

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

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

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

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

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

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

(c)

To determine

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

Answer to Problem 37E

The inflection points of $y=f\left(x\right)$ at $x=1$ and, $x=\frac{5}{3}$ .

Explanation of Solution

Given information :

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

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

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

Take factor common,

  $\begin{array}{c}{y^{\prime }}^{\prime }=\left(x-1\right)\left(2\left(x-2\right)+\left(x-1\right)\right)\\ =\left(x-1\right)\left(2x-4+x-1\right)\\ =\left(x-1\right)\left(3x-5\right)\end{array}$

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

Since, ${y^{\prime }}^{\prime }=\left(x-1\right)\left(3x-5\right)$ .

The second derivative is zero at $x=1$ and, $x=\frac{5}{3}$ .