Chapter 4.3, Problem 58E

(a)

To determine

To find:

The asymptote,maximum value,and inflextion points of $f$

Answer to Problem 58E

  $-x>0whenx<0$

So,function $f\left(x\right)$ is increasing on $\left(-\infty ,0\right)$

  $-x<0whenx>0$

So,function $f\left(x\right)$ is decreasing on $\left(0,\infty \right)$

Explanation of Solution

Given:

The family of bell shaped curves

  $y=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}$

Concept used:

The family of bell shaped curves

  $y=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\raisebox{1ex}{$-{\left(x-\mu \right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}$

Calculation:

The family of bell shaped curves

  $y=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\mathrm{.....................}\left(1\right)$

Since $x$ is squared and the same in both direction

  $\begin{array}{l}\underset{x\to \pm \infty }{\lim }y=\underset{x\to \pm \infty }{\lim }{e}^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\\ \underset{x\to \pm \infty }{\lim }y=0\end{array}$

Horizontal asymtote

  $y=0$

There are no vertical asymptote because the function is defined for all real numbers.

Differentiating equation (1) with respect to $x$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=0\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}{\left(\frac{-x^2}{2{\sigma }^2}\right)}^{\prime }\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{-2x^2}{2{\sigma }^2}\right)\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{-x^2}{{\sigma }^2}\right)\end{array}$

The only way this can equal $0$ is when $x=0$

  $-x>0whenx<0$

So,function $f\left(x\right)$ is increasing on $\left(-\infty ,0\right)$

  $-x<0whenx>0$

So,function $f\left(x\right)$ is decreasing on $\left(0,\infty \right)$

  $f\left(0\right)=1$ is the local and absolute max.

(b)

To determine

To find:

The value of $\sigma$ play in the shape of the curve.

Answer to Problem 58E

  $x=\pm \sigma$

Explanation of Solution

Given:

The family of bell shaped curves

  $y=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}$

Concept used:

The family of bell shaped curves

  $y=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\raisebox{1ex}{$-{\left(x-\mu \right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}$

Calculation:

The family of bell shaped curves

  $y=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\mathrm{.....................}\left(1\right)$

Since $x$ is squared and the same in both direction

  $\begin{array}{l}\underset{x\to \pm \infty }{\lim }y=\underset{x\to \pm \infty }{\lim }{e}^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\\ \underset{x\to \pm \infty }{\lim }y=0\end{array}$

Horizontal asymtote

  $y=0$

There are no vertical asymptote because the function is defined for all real numbers.

Differentiating equation (1) with respect to $x$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=0\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}{\left(\frac{-x^2}{2{\sigma }^2}\right)}^{\prime }\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{-2x^2}{2{\sigma }^2}\right)\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{-x^2}{{\sigma }^2}\right)\mathrm{.....................}\left(2\right)\end{array}$

Differentiating equation (2) with respect to $x$

  $\begin{array}{l}{f}^{″}\left(x\right)=0\\ f^{″}\left(x\right)={\left(e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\right)}^{\prime }\left(\frac{-x^2}{{\sigma }^2}\right)+e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}{\left(\frac{-x^2}{{\sigma }^2}\right)}^{\prime }\\ f^{″}\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{1}{{\sigma }^4}\right)\left(x^2-{\sigma }^2\right)\mathrm{.....................}\left(2\right)\\ 0=\left(x^2-{\sigma }^2\right)\\ x^2={\sigma }^2\\ x=\pm \sigma \end{array}$

(c)

To determine

To find:

The value of $\sigma$ play in the shape of the curve.

Answer to Problem 58E

Explanation of Solution

Given:

The family of bell shaped curves

  $y=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}$

Concept used:

The family of bell shaped curves

  $y=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\raisebox{1ex}{$-{\left(x-\mu \right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}$

Calculation:

The family of bell shaped curves

  $y=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\mathrm{.....................}\left(1\right)$

Since $x$ is squared and the same in both direction

  $\begin{array}{l}\underset{x\to \pm \infty }{\lim }y=\underset{x\to \pm \infty }{\lim }{e}^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\\ \underset{x\to \pm \infty }{\lim }y=0\end{array}$

Horizontal asymtote

  $y=0$

There are no vertical asymptote because the function is defined for all real numbers.

Differentiating equation (1) with respect to $x$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=0\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}{\left(\frac{-x^2}{2{\sigma }^2}\right)}^{\prime }\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{-2x^2}{2{\sigma }^2}\right)\\ f^{\prime }\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{-x^2}{{\sigma }^2}\right)\mathrm{.....................}\left(2\right)\end{array}$

Differentiating equation (2) with respect to $x$

  $\begin{array}{l}{f}^{″}\left(x\right)=0\\ f^{″}\left(x\right)={\left(e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\right)}^{\prime }\left(\frac{-x^2}{{\sigma }^2}\right)+e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}{\left(\frac{-x^2}{{\sigma }^2}\right)}^{\prime }\\ f^{″}\left(x\right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\left(\frac{1}{{\sigma }^4}\right)\left(x^2-{\sigma }^2\right)\mathrm{.....................}\left(2\right)\\ 0=\left(x^2-{\sigma }^2\right)\\ x^2={\sigma }^2\\ x=\pm \sigma \end{array}$

  $x<-\sigma then{x}^2>{\sigma }^{2}and\left(x^2-{\sigma }^2\right)>0$

So,CU on function $f\left(x\right)$ is $\left(-\infty ,-\sigma \right)$

  $-\sigma <x<\sigma whenx>0then{x}^2<{\sigma }^{2}and\left(x^2-{\sigma }^2\right)<0$

So,CD on function $f\left(x\right)$ is $\left(-\sigma ,\sigma \right)$

  $x>\sigma then{x}^2>{\sigma }^{2}and\left(x^2-{\sigma }^2\right)>0$

So,CU on function $f\left(x\right)$ is $\left(\sigma ,\infty \right)$

Inflection points are at

  $\begin{array}{l}f\left(\pm \sigma \right)=e^{\raisebox{1ex}{$-{\left(x\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2{\sigma }^2$}\right.}\\ f\left(\pm \sigma \right)=e^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ f\left(\pm \sigma \right)=\frac{1}{e^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$