Chapter 5.3, Problem 7E

a.

To determine

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

Answer to Problem 7E

The intervals are $\left(\frac{-7}{4},\infty \right)$

Explanation of Solution

Given Information:

The given equation is,

  $y=4x^3+21x^2+36x-20$

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 $y\text{'}\text{'}<0$ .

  $\begin{array}{l}y=4x^3+21x^2+36x-20\\ y^1=12x^2+42x+36\left(\text{first order derivative}\right)\\ y^{11}=24x+42\left(\text{second order derivative}\right)\\ \text{and,}\\ y^{11}=24x+42=0\end{array}$

The value of x must be,

  $\begin{array}{l}x=\frac{-42}{24}=\frac{-7}{4}\left[\text{gcd of }42\text{and 24 is 6}\right]\\ y^{11}\left(-2\right)=24\left(-2\right)+42=-6<0\\ y^{11}\left(0\right)=24\left(0\right)+42=42>0\end{array}$

The $y$ is then concave up for $x>\frac{-7}{4}$

Hence,

The intervals are $\left(\frac{-7}{4},\infty \right)$

b.

To determine

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

Answer to Problem 7E

The intervals are $\left(-\infty ,\frac{-7}{4}\right)$

Explanation of Solution

Given Information:

Given equation is,

  $y=4x^3+21x^2+36x-20$

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 $y\text{'}\text{'}<0$ .

  $\begin{array}{l}y=4x^3+21x^2+36x-20\\ y^1=12x^2+42x+36\left(\text{first order derivative}\right)\\ y^{11}=24x+42\left(\text{second order derivative}\right)\\ \text{and,}\\ y^{11}=24x+42=0\end{array}$

The value of x must be,

  $\begin{array}{l}x=\frac{-42}{24}=\frac{-7}{4}\left[\text{gcd of }42\text{and 24 is 6}\right]\\ y^{11}\left(-2\right)=24\left(-2\right)+42=-6<0\\ y^{11}\left(0\right)=24\left(0\right)+42=42>0\end{array}$

The $y$ is then concave up for $x<\frac{-7}{4}$

Hence,

The intervals are $\left(-\infty ,\frac{-7}{4}\right)$