Question

In Exercises 27 and 28 , find an equation for the line tangent to the curve at the given point. $$y=\frac{x^{3}+1}{2 x}, x=1$$

Short answer

The equation of the line tangent to the curve \(y=\frac{x^{3}+1}{2 x}\) at the point \(x=1\) is \(y - 1 = 1/2 (x - 1)\).

Step-by-step solution

Compute The derivative

The derivative of the function \(y=\frac{x^{3}+1}{2 x}\) can be computed using quotient rule. The quotient rule states that the derivative of a quotient of two functions is given by \((fu'-uf')/ v^2\). Here, \(u=x^3+1\) and \(v=2x\). Thus, \(u'=3x^2\) and \(v'=2\). Substituting these values into the quotient rule gives: \(y'=\frac{(2x)(3x^2)-(x^3+1)(2)}{(2x)^2}\).

Substitute x=1 Into The Determined Derivative

Substitute \(x=1\) into the equation \(y'=\frac{(2x)(3x^2)-(x^3+1)(2)}{(2x)^2}\) to determine the slope of the tangent at that point. After substitution we get \(y'=\frac{1}{2}\). This means that the slope of the tangent line at the point where \(x=1\) is \(1/2\).

Find The Corresponding y Value At x=1

Substitute \(x=1\) into the original equation \(y=\frac{x^{3}+1}{2 x}\) to find the corresponding y value. After substitution we get \(y=1\). So the point of tangency is \(1,1\).

Find The Equation of The Tangent Line

Finally, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form of an equation of a line is given by \(y - y1 = m(x – x1)\), where \((x1, y1)\) is a known point on the line and \(m\) is the slope of the line. Plugging in \(m=1/2\), \(x1=1\) and \(y1=1\), we get the equation \(y - 1 = 1/2 (x - 1)\).