Chapter 3.3, Problem 26E

To determine

The equation for the line tangent to the curve at the given point.

Answer to Problem 26E

The required equation is $y=\frac{1}{2}x-\frac{1}{2}$ .

Explanation of Solution

Given information:

The curve: $y=\frac{x^3+1}{2x}$

The point: $x=1$

Formula used:

Differentiation formula:

  $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

Point-slope form of the equation of a line:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The curve is $y=\frac{x^3+1}{2x}$ .

And, the point is $x=1$ .

Substitute 1 for x and 1 for y in the given curve.

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

So, the point through which the tangent line passes is $\left(1,1\right)$ .

To find the slope of the tangent line to the given curve, differentiate the above curve.

Use the differentiation formula $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$ .

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

Simplify the expression.

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

Substitute 1 for x in the above derivative.

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

Thus, the slope of the tangent line is $\frac{1}{2}$ .

Now, to find the equation of the tangent line use the point-slope form of an equation of a line $y-y_1=m\left(x-x_1\right)$ .

Substitute 1 for $x_1$ , 1 for $y_1$ and $\frac{1}{2}$ for m in the above formula.

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

Hence, the required equation is $y=\frac{1}{2}x-\frac{1}{2}$ .