Chapter 3.2, Problem 48E

(a)

To determine

To show : when $x=2$ , the curve possesses more than one value of $y$ .

Explanation of Solution

Given information :

The curve is $y^3-xy=-1$ .

Substitute $x=2$ in $y^3-xy=-1$ .

  $\begin{array}{c}{y}^3-2y=-1\\ y^3-2y+1=0\\ y^3-y^2+y^2-y-y+1=0\\ y^2\left(y-1\right)+y\left(y-1\right)-1\left(y-1\right)=0\\ \left(y^2+y-1\right)\left(y-1\right)=0\\ \left(y^2+y-1\right)=0,\left(y-1\right)=0\\ y=-1.618,0.618,1\end{array}$

The three values are possible when $x=2$ substitute in curve $y^3-xy=-1$ .

Hence, the relation $y^3-xy=-1$ cannot be a function of $x$ .

(b)

To determine

To find : the values of function $f^{\prime }\left(2\right)$ and $f^{″}\left(2\right)$ .

Answer to Problem 48E

  $f^{\prime }\left(2\right)=\frac{1}{2}$ and $f^{″}\left(2\right)=-\frac{1}{4}$ .

Explanation of Solution

Given information :

The curve is $y^3-xy=-1$ .

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

The graph of curve near $\left(2,1\right)$ is

  

Take the curve $y^3-xy=-1$ .

  $y\left(y^2-x\right)=-1$

Either $y=1$ or $y^2-x=-1$ .

So, $y=\sqrt{x-1}\Rightarrow f\left(x\right)=\sqrt{x-1}$

Differentiate $f\left(x\right)=\sqrt{x-1}$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\sqrt{x-1}\\ =\frac{1}{2\sqrt{x-1}}\frac{d}{dx}\left(x-1\right)\\ =\frac{1}{2\sqrt{x-1}}\end{array}$

Substitute $x=2$ in $f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x-1}}$ .

  $\begin{array}{c}{f}^{\prime }\left(2\right)=\frac{1}{2\sqrt{2-1}}\\ =\frac{1}{2\sqrt{1}}\\ =\frac{1}{2}\end{array}$

Differentiate $f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x-1}}$ with respect to $x$ ,

  $\begin{array}{c}{f}^{″}\left(x\right)=\frac{d}{dx}\left(\frac{1}{2\sqrt{x-1}}\right)\\ =\frac{1}{2}\frac{d}{dx}{\left(x-1\right)}^{-1/2}\\ =\frac{1}{2}\cdot \left(\frac{-1}{2}\right){\left(x-1\right)}^{-3/2}\\ =-\frac{1}{4}{\left(x-1\right)}^{-3/2}\end{array}$

Substitute $x=2$ in $f^{″}\left(x\right)=-\frac{1}{4}{\left(x-1\right)}^{-3/2}$ .

  $\begin{array}{c}{f}^{″}\left(2\right)=-\frac{1}{4}{\left(2-1\right)}^{-3/2}\\ =-\frac{1}{4}{\left(1\right)}^{-3/2}\\ =-\frac{1}{4}\end{array}$

Hence, $f^{\prime }\left(2\right)=\frac{1}{2}$ and $f^{″}\left(2\right)=-\frac{1}{4}$ .