Chapter 5.3, Problem 11E

a.

To determine

To determine the intervals on which the graph of the function is concave up.

Answer to Problem 11E

  $y^{″}=0$ is concave up.

Explanation of Solution

Given Information:

Given equation is,

  $y=\{\begin{array}{l}2x,x<1\\ 2-x^2,x\ge 1\end{array}$

Calculation:

The graph of a differentiable function $y=f\left(x\right)$ is concave up on an open interval $I$ if $y\text{'}$ is increasing on $I$ .

Where if a function $y=f\left(x\right)$ has a second derivative then it can conclude that $y\text{'}$ increases if $y\text{'}\text{'}>0$ and decreases if $y\text{'}\text{'}<0$ .

  $\begin{array}{l}{y}^{\prime }=2,x<1\\ y^{\prime }=-2x,x\ge 1\\ y^{″}=0,x<1\\ y^{″}=-2,x\ge 1\end{array}$

Because $y^{″}=0$ for $x<1$ the concavity test for concave up is failed when $x<1$ .

Hence,

  $y^{″}=0$ is concave up,

b.

To determine

To determine the intervals on which the graph of the function is concave down.

Answer to Problem 11E

Function is concave down at interval $\left(1,\infty \right)$ .

Explanation of Solution

Given Information:

Given equation is,

  $y=\{\begin{array}{l}2x,x<1\\ 2-x^2,x\ge 1\end{array}$

Calculation:

The graph of a differentiable function $y=f\left(x\right)$ is concave down on an open interval $I$ if $y\text{'}$ is increasing on $I$ .

Where if a function $y=f\left(x\right)$ has a second derivative then it can conclude that $y\text{'}$ increases if $y\text{'}\text{'}>0$ and decreases if $y\text{'}\text{'}<0$ .

  $\begin{array}{l}{y}^{\prime }=2,x<1\\ y^{\prime }=-2x,x\ge 1\\ y^{″}=0,x<1\\ y^{″}=-2,x\ge 1\end{array}$

Since, $y^{″}=-2$ or $y\text{'}\text{'}<0$ for $x\ge 1$ ,

Hence,

Function is concave down at interval $\left(1,\infty \right)$ .