Single Variable Calculus, Chapter 2, 2.1, Section 2.1, Problem 3

Given the curve $\displaystyle y= \frac{x}{1+x}$ and the point $J \left(\displaystyle 1, \frac{1}{2}\right)$ that lies on the curve.

a. Use your calculator to find the slope of the secant line $JK$ (correct to six decimal places) for the following values of $x$ if $K$ is the point $\displaystyle \left( x, \frac{x}{1+x} \right)$.


$
\begin{equation}
\begin{aligned}

&\text{ (i) } 0.5 && \text{ (v) } 1.5 \\
&\text{ (ii) } 0.9 && \text{(vi) } 1.1 \\
&\text{ (iii) }0.99 && \text{(vii) } 1.01\\
&\text{ (iv) } 0.999 && \text{(viii) } 1.001
\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

\begin{array}{|c|c|c|c|c|c|}
\hline\\
& K_x(x) & K_y \left(\frac{x}{1 + x}\right) & J_x & J_y & \text{Slope of secant line } JK = \left(\frac{J_y - K_y}{J_x - K_x}\right) \\
\hline\\
(i) & 0.5 & \frac{1}{3} & 1 & \frac{1}{2} & 0.333333 \\
\hline\\
(ii) & 0.9 & \frac{9}{19} & 1 & \frac{1}{2}& 0.263158 \\
\hline\\
(iii) & 0.99 & \frac{99}{199} & 1 & \frac{1}{2} & 0.251256 \\
\hline\\
(iv) & 0.999 & \frac{999}{1999} & 1 & \frac{1}{2} & 0.250125 \\
\hline\\
(v) & 1.5 & \frac{3}{5} & 1 & \frac{1}{2} & 0.20 \\
\hline\\
(vi) & 1.1 & \frac{11}{21} & 1 & \frac{1}{2} & 0.238095 \\
\hline\\
(vii) & 1.01 & \frac{101}{201} & 1 & \frac{1}{2} & 0.248756 \\
\hline\\
(viii) & 1.001 & \frac{1001}{2001} & 1 & \frac{1}{2} & 0.249875\\
\hline
\end{array}
\end{aligned}
\end{equation}
$


b. Guess the value of the slope of the tangent line to the curve at $J \left(\displaystyle 1, \frac{1}{2}\right)$ using the results in part (a).

Based from the values we obtain from the table, the slope of the tangent line seems to have a value of $\displaystyle \frac{1}{4}$

c. Find an equation of the tangent line to the curve at $J\left(\displaystyle 1, \frac{1}{2}\right)$ using the slope in part (b).

Using point slope form:

$\displaystyle y - \frac{1}{2} = \frac{1}{4}(x - 1)$

$\displaystyle y = \frac{x}{4} + \frac{1}{4}$