Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 4

a.) Determine the slope of the tangent line to the curve $y = x - x^3$ at the point $P (1, 0)$

$(i) \text{ Using the definition (Slope of the tangent line)}$


$
\begin{equation}
\begin{aligned}

\displaystyle m &= \lim \limits_{x \to 1} \frac{f(x) - f(1)}{x - 1} && \\
\\
\displaystyle m &= \lim \limits_{x \to 1} \frac{x - x^3 - [1 - (1)^3]}{x - 1}
&& \text{ Substitute value of $a$ and $x$}\\
\\
\displaystyle m &= \lim \limits_{x \to 1} \frac{-x(x + 1) \cancel{(x - 1)}}{\cancel{x - 1}}
&& \text{ Cancel out like terms}\\
\\
\displaystyle m &= \lim \limits_{x \to 1} [- x(x + 1)] = \lim \limits_{x \to 1} ( -x^2-x) = -(1)^2 - (1) = -2
&& \text{ Evaluate the limit}\\

\end{aligned}
\end{equation}
$


Therefore,
The slope of the tangent line is $m = -2$
$(ii)\text{ Using the equation}$

$\displaystyle m = \lim \limits_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Let $f(x) = x - x^3$ So the slope of the tangent line at $(1, 0)$ is


$
\begin{equation}
\begin{aligned}

\displaystyle m =& \lim \limits_{h \to 0} \frac{f( 1 + h) - f(1)}{h} && \\
\\
\displaystyle m =& \lim \limits_{h \to 0} \frac{1 + h - (1 + h)^3 - [1 - (1)^3]}{h}
&& \text{ Substitute value of $a$}\\
\\
\displaystyle m =& \lim \limits_{h \to 0} \frac{1 + h - (1 + 3h^2 + 3h + h^3)}{h}
&& \text{ Expand and simplify }\\
\\
\displaystyle m =& \lim \limits_{h \to 0} \frac{-2h - 3h^2 - h^3}{h}
&& \text{ Factor the numerator}\\
\\
\displaystyle m =& \lim \limits_{h \to 0} \frac{\cancel{h}(-2 - 3h -h^2)}{\cancel{h}} = \lim \limits_{h \to 0} (-2-3h-h^)
&& \text{ Cancel out like terms}
\\
m =& -2-3(0) - (0)^2 = -2

\end{aligned}
\end{equation}
$


Therefore,
The slope of the tangent line is $m = -2$
b.) Write an expression of the tangent line in part (a)

Using the point slope form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m ( x - x_1)\\
\\
y - 0 =& -2 ( x - 1)
&& \text{ Substitute value of $x, y$ and $m$ and simplify}\\
\end{aligned}
\end{equation}
$

Therefore,
The equation of the tangent line at $(1,0)$ is $y = -2x + 2$

c.) Draw a graph of the curve and the tangent line in successively smaller viewing rectangles centered at $(1, 0)$ until the curve and the line appear to coincide.