Chapter 2.3, Problem 28E

To determine

To calculate: The equations of tangent to the given curve.

Answer to Problem 28E

The equation of tangent line is $y=-\frac{3}{2}\left(x+1\right)$ .

Explanation of Solution

Given information:

The equation of the curve is $y=\frac{x^4+2}{x^2}$ , $x=-1$

Concept used:

The formulae used are $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ , quotient rule is $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v.\frac{d}{dx}u-u.\frac{d}{dx}v}{v^2}$ .

Calculation:

Slope of tangent line.

  

Solve further.

  $\begin{array}{c}m=\frac{6x^3-2x^3-2}{4x^2}\\ =\frac{4x^3-2}{4x^2}\\ =\frac{2\left(2x^3-1\right)}{4x^2}\\ =\frac{2x^3-1}{2x^2}\end{array}$

Solve further by substituting $-1$ for $x$ in $m=\frac{2x^3-1}{2x^2}$

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

The slope of tangent line is $-\frac{3}{2}$

Substitute for $x$ in $y=\frac{x^3+1}{2x}$ for the value of $y_1$ .

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

So, the value of $\left(x_1,y_1\right)=\left(-1,0\right)$

Substitute the values in the standard equation of line $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{c}y-0=-\frac{3}{2}\left(x+1\right)\\ y=1-\frac{3}{2}\left(x+1\right)\end{array}$

The equation of tangent line is $y=-\frac{3}{2}\left(x+1\right)$.

Conclusion: The equation of tangent line is $y=-\frac{3}{2}\left(x+1\right)$ .