Chapter 2.3, Problem 27E

To determine

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

Answer to Problem 27E

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

Explanation of Solution

Given information:

The equation of the curve is $y=\frac{x^3+1}{2x}$ , $x=1$

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.

  $\begin{array}{c}m=\frac{dy}{dx}\\ =\frac{d}{dx}\left(\frac{x^3+1}{2x}\right)\\ =\frac{2x.\frac{d}{dx}\left(x^3+1\right)-\left(x^3+1\right).\frac{d}{dx}\left(2x\right)}{{\left(2x\right)}^2}\\ =\frac{2x.\left(3x^2\right)-\left(x^3+1\right).2}{4x^2}\end{array}$

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 $m=\frac{2x^3-1}{2x^2}$

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

The slope of tangent line is $\frac{1}{2}$

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

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

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

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

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

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

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