Chapter 4, Problem 26RE

(a)

To determine

To find:

The number of g have an extreme value.

Answer to Problem 26RE

  $g^{″}=0$

  $\text{when}x=0$

Explanation of Solution

Given:

The function

  $g\left(x\right)=f\left(x^2\right)$

  $f^{″}\left(x\right)<0\text{for}x<0\text{and}{f}^{″}\left(x\right)>0\text{for}x>0$

Concept used:

The left derivative and right derivative of a function $f$ at a point $x=c$ are , equal

  $\begin{array}{l}{f}_{-}\text{'}\left(c\right)=\underset{h\to 0^{-}}{\lim }\frac{f\left(c-h\right)-f\left(c\right)}{-h}\\ f_{+}\text{'}\left(c\right)=\underset{h\to 0^{+}}{\lim }\frac{f\left(c+h\right)-f\left(c\right)}{h}\end{array}$

Calculation:

  $g\left(x\right)=f\left(x^2\right)\mathrm{..................}\left(1\right)$

Differentiating equation (1) with respect to $x$

  $g^{\prime }\left(x\right)=2x{f}^{\prime }\left(x^2\right)\mathrm{.....................}\left(2\right)$

Differentiating equation (1) with respect to $x$ by the product and chain rules

  $\begin{array}{l}{g}^{\prime }\left(x\right)=2x{f}^{\prime }\left(x^2\right)\\ g^{″}\left(x\right)={\left(2x\right)}^{\prime }{f}^{\prime }\left(x^2\right)+2x{\left[f^{\prime }\left(x^2\right)\right]}^{\prime }\\ g^{″}\left(x\right)=2f^{\prime }\left(x^2\right)+2x\left[2x{f}^{″}\left(x^2\right)\right]\\ g^{″}\left(x\right)=2f^{\prime }\left(x^2\right)+4x^2\left[f^{″}\left(x^2\right)\right]\end{array}$

  $g^{″}=0$

  $\text{when}x=0$

(b)

To determine

To find:

The number of g have the concavity .

Answer to Problem 26RE

  $g$ is concave up for all real numbers.

Explanation of Solution

Given:

The function

  $g\left(x\right)=f\left(x^2\right)$

  $f^{″}\left(x\right)<0\text{for}x<0\text{and}{f}^{″}\left(x\right)>0\text{for}x>0$

Concept used:

The left derivative and right derivative of a function $f$ at a point $x=c$ are , equal

  $\begin{array}{l}{f}_{-}\text{'}\left(c\right)=\underset{h\to 0^{-}}{\lim }\frac{f\left(c-h\right)-f\left(c\right)}{-h}\\ f_{+}\text{'}\left(c\right)=\underset{h\to 0^{+}}{\lim }\frac{f\left(c+h\right)-f\left(c\right)}{h}\end{array}$

Calculation:

  $g\left(x\right)=f\left(x^2\right)\mathrm{..................}\left(1\right)$

Differentiating equation (1) with respect to $x$

  $g^{\prime }\left(x\right)=2x{f}^{\prime }\left(x^2\right)\mathrm{.....................}\left(2\right)$

Differentiating equation (1) with respect to $x$ by the product and chain rules

  $\begin{array}{l}{g}^{\prime }\left(x\right)=2x{f}^{\prime }\left(x^2\right)\\ g^{″}\left(x\right)={\left(2x\right)}^{\prime }{f}^{\prime }\left(x^2\right)+2x{\left[f^{\prime }\left(x^2\right)\right]}^{\prime }\\ g^{″}\left(x\right)=2f^{\prime }\left(x^2\right)+2x\left[2x{f}^{″}\left(x^2\right)\right]\\ g^{″}\left(x\right)=2f^{\prime }\left(x^2\right)+4x^2\left[f^{″}\left(x^2\right)\right]\end{array}$

Given that

  $f^{″}\left(x\right)<0\text{for}x<0\text{and}{f}^{″}\left(x\right)>0\text{for}x>0$

  $x^2$ is positive for $x\ne 0,\text{so}{f}^{″}\left(x^2\right)>0\text{for}x\ne 0$

  $g^{″}=0$ $;x=0$

Since $g^{″}$ is positive on either side that $g$ is concave up for all real numbers.