Chapter 4.2, Problem 48E

(a)

To determine

To show: The relation cannot be a function of x by showing that there is more than one possible y- value when $x=2$ .

Answer to Problem 48E

The relation cannot be a function of x.

Explanation of Solution

Given Information: The given equation is $y^3-xy=-1$

Calculation:

Substitute $x=2$ in the equation $y^3-xy=-1$ ,

  $\begin{array}{c}{y}^3-xy=-1\\ y^3-2y=-1\\ y^3-2y+1=0\end{array}$

Find the derivative and equate it to 0 to obtain the values of y,

  $\begin{array}{c}\frac{d}{dy}\left(y^3-2y+1\right)=0\\ 3y^2-2=0\\ 3y^2=2\\ y^2=\frac{2}{3}\\ y=\pm \sqrt{\frac{2}{3}}\end{array}$

Thus, the relation cannot be a function of x.

(b)

To determine

To calculate: $f\text{'}\left(2\right)\text{ and }f\text{'}\text{'}\left(2\right)$ .

Answer to Problem 48E

The value of $f\text{'}\left(2\right)=1\text{ and }f\text{'}\text{'}\left(2\right)=-4$ .

Explanation of Solution

Given Information: On a small enough square with center $\left(2,1\right)$ , the part of the graph of the relation within the square defines a function $y=f\left(x\right)$ .

Calculation:

The derivative of the curve is,

  $\begin{array}{c}{y}^3-xy=-1\\ 3y^2\frac{dy}{dx}-\left(x\frac{dy}{dx}+y\right)=\frac{d}{dx}\left(-1\right)\\ 3y^2\frac{dy}{dx}-x\frac{dy}{dx}-y=0\\ \left(3y^2-x\right)\frac{dy}{dx}-y=0\\ \left(3y^2-x\right)\frac{dy}{dx}=y\\ \frac{dy}{dx}=\frac{y}{\left(3y^2-x\right)}\\ f\text{'}\left(x\right)=\frac{y}{\left(3y^2-x\right)}\end{array}$

The value of the derivative at $\left(2,1\right)$ is,

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

The value of the second derivative at $\left(2,1\right)$ is,

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{y}{\left(3y^2-x\right)}\\ f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{y}{3y^2-x}\right)\\ =\frac{\left(3y^2-x\right)\frac{dy}{dx}-y\frac{d}{dx}\left(3y^2-x\right)}{{\left(3y^2-x\right)}^2}\\ =\frac{\left(3y^2-x\right)\frac{dy}{dx}-y\left(6y\frac{dy}{dx}-1\right)}{{\left(3y^2-x\right)}^2}\\ =\frac{3y^2\frac{dy}{dx}-x\frac{dy}{dx}-6y^2\frac{dy}{dx}+y}{{\left(3y^2-x\right)}^2}\\ =\frac{\left(-3y^2-x\right)\left(\frac{dy}{dx}\right)+y}{{\left(3y^2-x\right)}^2}\\ =\frac{\left(-3{\left(1\right)}^2-2\right)\times 1+1}{{\left(3{\left(1\right)}^2-2\right)}^2}\left(\text{From part (a) }\frac{dy}{dx}=1\right)\\ =\frac{-3-2+1}{{\left(3-2\right)}^2}\\ =\frac{-4}{1}\\ =-4\end{array}$