Chapter 2.3, Problem 4QQ

a.

To determine

To calculate: The points of horizontal tangents.

Answer to Problem 4QQ

The points are $0,x=\sqrt{2},-\sqrt{2}$ .

Explanation of Solution

Given information:

The function is. $f\left(x\right)=x^4-4x^2$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Calculation:

The function is $f\left(x\right)=x^4-4x^2$ .

Differentiate the function with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x^4-4x^2\right)\\ =\frac{d}{dx}\left(x^4\right)-4\frac{d}{dx}\left(x^2\right)\\ =4x^3-8x\\ \end{array}$

Equate the derivative to zero for the point of horizontal tangents.

  $\begin{array}{l}4x^3-8x=0\\ 4x\left(x^2-2\right)=0\\ 4x=0,x^2-2=0\\ x=0,x=\sqrt{2},-\sqrt{2}\end{array}$

Substitute $0$ for $x$ in $y=x^4-4x^2$ .

  $\begin{array}{c}y={\left(0\right)}^4-4{\left(0\right)}^2\\ y=0\end{array}$

Substitute $\sqrt{2}$ for $x$ in $y=x^4-4x^2$ .

  $\begin{array}{c}y={\left(\sqrt{2}\right)}^4-4{\left(\sqrt{2}\right)}^2\\ y=4-8\\ y=-4\end{array}$

Substitute $-\sqrt{2}$ for $x$ in $y=x^4-4x^2$ .

  $\begin{array}{c}y={\left(-\sqrt{2}\right)}^4-4{\left(-\sqrt{2}\right)}^2\\ y=4-8\\ y=-4\end{array}$

Conclusion: The equation of tangents are $y=0$ , $y=-4$ .

b.

To determine

To calculate: The equation of tangent at $x=1$ .

Answer to Problem 4QQ

The equation of tangent is $y=-3-4\left(x-1\right)$ .

Explanation of Solution

Given information:

The function is. $f\left(x\right)=x^4-4x^2$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

The function is $f\left(x\right)=x^4-4x^2$ .

Differentiate the function with respect to $x$

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x^4-4x^2\right)\\ =\frac{d}{dx}\left(x^4\right)-4\frac{d}{dx}\left(x^2\right)\\ =4x^3-8x\end{array}$

Substitute $1$ for $x$ in $m=4x^3-8x$ .

  $\begin{array}{c}m=4{\left(1\right)}^3-8\left(1\right)\\ =4-8\\ =-4\end{array}$

Substitute $1$ for $x$ in $y=x^4-4x^2$ .

  $\begin{array}{c}y={\left(1\right)}^4-4{\left(1\right)}^2\\ =1-4\\ y=-3\end{array}$

Substitute the values in the formula $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-\left(-3\right)=-4\left(x-1\right)\\ y=-3-4\left(x-1\right)\end{array}$

Conclusion: The equation of tangent is $y=-3-4\left(x-1\right)$ .

c.

To determine

To calculate: The equation of normal line at $x=1$ .

Answer to Problem 4QQ

The equation of normal is $y=-3-\frac{1}{4}\left(x-1\right)$ .

Explanation of Solution

Given information:

The function is. $f\left(x\right)=x^4-4x^2$

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

The function is $f\left(x\right)=x^4-4x^2$ .

Differentiate the function with respect to $x$

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x^4-4x^2\right)\\ =\frac{d}{dx}\left(x^4\right)-4\frac{d}{dx}\left(x^2\right)\\ =4x^3-8x\end{array}$

Substitute $1$ for $x$ in $m=4x^3-8x$ .

  $\begin{array}{c}m=4{\left(1\right)}^3-8\left(1\right)\\ =4-8\\ =-4\end{array}$

Use the formula $m_1\times m_2=-1$ .

  $\begin{array}{c}{m}_1\times m_2=-1\\ m_2=\frac{-1}{-4}\\ m_2=\frac{1}{4}\end{array}$

The slope of normal is $\frac{1}{4}$ .

Substitute $1$ for $x$ in $y=x^4-4x^2$ .

  $\begin{array}{c}y={\left(1\right)}^4-4{\left(1\right)}^2\\ =1-4\\ y=-3\end{array}$

Substitute the values in the formula $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-\left(-3\right)=\frac{1}{4}\left(x-1\right)\\ y=-3-\frac{1}{4}\left(x-1\right)\end{array}$

Conclusion: The equation of normal is $y=-3-\frac{1}{4}\left(x-1\right)$ .