Chapter 12.3, Problem 57E

To determine

ToFind:The equation of the line that is tangent to the graph of $f\left(x\right)=-\frac{1}{2}{x}^3$ and parallel to the line $6x+y+4=0$

Answer to Problem 57E

The equation of the tangent line to the graph of $f\left(x\right)=-\frac{1}{2}{x}^3$ and parallel to the line $6x+y+4=0$ is $y=-6x-8$ and $y=-6x+8$

Explanation of Solution

Given:

The function $f\left(x\right)=-\frac{1}{2}{x}^3$ and the line $6x+y+4=0$

Concept Used:

The slope $m$ of the tangent lineto the graph of $f$ at $\left(x,f\left(x\right)\right)$ , and is given by $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

  ${\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

Calculation:

Forthe function $f\left(x\right)=-\frac{1}{2}{x}^3$

The slope $m$ of the tangent line to the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is given by

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\left(-\frac{1}{2}{\left(x+h\right)}^3\right)-\left(-\frac{1}{2}{x}^3\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\left(-\frac{1}{2}\left(x^3+h^3+3x^{2}h+3x{h}^2\right)\right)-\left(-\frac{1}{2}{x}^3\right)}{h}\right){\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{-\frac{1}{2}{x}^3-\frac{1}{2}{h}^3-\frac{3}{2}{x}^{2}h-\frac{3}{2}x{h}^2+\frac{1}{2}{x}^3}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{-\frac{1}{2}{h}^3-\frac{3}{2}{x}^{2}h-\frac{3}{2}x{h}^2}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{-h^3-3x^{2}h-3x{h}^2}{2h}\right)\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{h\left(-h^2-3x^2-3xh\right)}{2h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{-h^2-3x^2-3xh}{2}\right)\\ =\frac{-{\left(0\right)}^2-3x^2-3x\left(0\right)}{2}\text{ Substituting limit}\\ =\frac{-3x^2}{2}\end{array}$

Consider the line $6x+y+4=0$

Writing the line $6x+y+4=0$ into slope intercept form

  $y=-6x-4$

The slope of the line $6x+y+4=0$ is $-6$

Since, the tangent line is parallel to the line $6x+y+4=0$ . So, their slopes will be equal.

  $\begin{array}{c}\frac{-3x^2}{2}=-6\\ -3x^2=-12\\ x^2=\frac{-12}{-3}\\ x^2=4\end{array}$

  $x=-2$

and

  $x=2$

For $x=-2$

Substituting in the function $y=f\left(x\right)=-\frac{1}{2}{x}^3$

  $\begin{array}{c}y=-\frac{1}{2}{\left(-2\right)}^3\\ =-\frac{\left(-8\right)}{2}\\ =4\end{array}$

For $x=2$

Substituting in the function $y=f\left(x\right)=-\frac{1}{2}{x}^3$

  $\begin{array}{c}y=-\frac{1}{2}{\left(2\right)}^3\\ =-\frac{\left(8\right)}{2}\\ =-4\end{array}$

So, the points on the tangent line are $\left(x,y\right)=\left(-2,4\right)$ and $\left(x,y\right)=\left(2,-4\right)$

Also, the slope of tangent line is $m=-6$

The equation of the tangent line at the point $\left(a,b\right)$ is given by

  $y-b=m\left(x-a\right)$

At $\left(x,y\right)=\left(-2,4\right)$

  $\begin{array}{c}\left(y-4\right)=-6\left(x-\left(-2\right)\right)\\ y-4=-6x-12\\ y=-6x-12+4\\ y=-6x-8\end{array}$

At $\left(x,y\right)=\left(2,-4\right)$

  $\begin{array}{c}\left(y-\left(-4\right)\right)=-6\left(x-2\right)\\ y+4=-6x+12\\ y=-6x+12-4\\ y=-6x+8\end{array}$

Therefore, the equation of the tangent line to the graph of $f\left(x\right)=-\frac{1}{2}{x}^3$ and parallel to the line $6x+y+4=0$ is $y=-6x-8$ and $y=-6x+8$

Conclusion:

Theequation of the tangent line to the graph of $f\left(x\right)=-\frac{1}{2}{x}^3$ and parallel to the line $6x+y+4=0$ is $y=-6x-8$ and $y=-6x+8$