Chapter 2.3, Problem 58E

a.

To determine

To explain why the local maxima and minima of $f$ and $g$ occur at the same value of $x$ .

Answer to Problem 58E

The local maxima and minima of $f$ and $g$ occur at the same value of $x$ because it is impossible for $g$ to have local maxima and minima at $x=a$ if $f$ do not have local maxima and minima at the same point $x=a$ .

Explanation of Solution

Given information:

the function is $g\left(x\right)=\sqrt{f\left(x\right)}$

The local maxima of $g$ will occur at $x=a$ if and only if $g\left(a\right)\ge g\left(x\right)$ since, $g\left(x\right)=\sqrt{f\left(x\right)}$ so, conversely, $g\left(a\right)\ge g\left(x\right)$ will be true if and only if $f\left(a\right)\ge f\left(x\right)$ .

Similarly,

The local minima of $g$ will occur at $x=a$ if and only if $g\left(a\right)\le g\left(x\right)$ since, $g\left(x\right)=\sqrt{f\left(x\right)}$ so, conversely, $g\left(a\right)\le g\left(x\right)$ will be true if and only if $f\left(a\right)\le f\left(x\right)$ .

Hence,

The local maxima and minima of $f$ and $g$ occur at the same value of $x$ because it is impossible for $g$ to have local maxima and minima at $x=a$ if $f$ do not have local maxima and minima at the same point $x=a$ .

b.

To determine

To express $g$ as a function of $x$ .

Answer to Problem 58E

$g$ as a function of $x$ is $g\left(x\right)=\sqrt{x^4+x^2-6x+9}$

Explanation of Solution

Given information:

  $g\left(x\right)$ is the distance between the point $\left(3,0\right)$ and the point $\left(x,x^2\right)$ on the graph of the parabola $y=x^2$ .

Formula used:

Distance between two point $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is defined by,

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

  $g\left(x\right)$ is the distance between the point $\left(3,0\right)$ and the point $\left(x,x^2\right)$ on the graph of the parabola $y=x^2$ .

Therefore,

Use distance formula,

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ g\left(x\right)=\sqrt{{\left(x-3\right)}^2+{\left(x^2-0\right)}^2}\\ g\left(x\right)=\sqrt{x^2-6x+9+x^4}\\ g\left(x\right)=\sqrt{x^4+x^2-6x+9}\end{array}$

Hence,

  $g$ as a function of $x$ is $g\left(x\right)=\sqrt{x^4+x^2-6x+9}$

c.

To determine

To find the minimum value of the function $g$ .

Answer to Problem 58E

The minimum value of the function $g$ is 2.236 at $x=1$ .

Explanation of Solution

Given information:

  $g\left(x\right)$ is the distance between the point $\left(3,0\right)$ and the point $\left(x,x^2\right)$ on the graph of the parabola $y=x^2$ .

Calculation:

The function is $g\left(x\right)=\sqrt{x^4+x^2-6x+9}$ from part (b).

Use the principle describe in part (a) that the local maxima and minima of $f$ and $g$ occur at the same value of $x$ .

Compare $g\left(x\right)=\sqrt{x^4+x^2-6x+9}$ with $g\left(x\right)=\sqrt{f\left(x\right)}$ ,

  $f\left(x\right)=x^4+x^2-6x+9$

The graph of $f$ is shown below:

  

There appears to be one local minimum at $x=1$ .

Hence,

Using maximum command, the local minimum value of $y$ is 5 and this value occurs when $x$ is 1.

Since, the local maxima and minima of $f$ and $g$ occur at the same value of $x$ .

Hence, localminima of $g$ occur at the same value of $x$ that is $x=1$ .

So, the minimum value of $g$ is obtained as:

  $\begin{array}{l}g\left(1\right)=\sqrt{f\left(1\right)}\\ g\left(1\right)=\sqrt{5}\\ g\left(1\right)=2.236\end{array}$

Hence,

The minimum value of the function $g$ is 2.236 at $x=1$ .