Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 22

Suppose that $g(x) = 1 - x^3$, find $g'(0)$ and use it to find an equation of the tangent line to the curve $y = 1 - x^3$ at the point $(0, 2)$

Using the definition of the derivative of a function $g$ at a number $a$, denoted by $g'(a)$, is

$\qquad \displaystyle g'(a) = \lim \limits_{h \to 0} \frac{g(a + h ) - g(a)}{h}$

We have,


$
\begin{equation}
\begin{aligned}

\qquad g'(a) =& \lim \limits_{h \to 0} \frac{1 - (a + h)^3 - (1 - a^3)}{h}
&& \text{Substitute $g(a + h)$ and $g(a)$}\\
\\
\qquad g'(a) =& \lim \limits_{h \to 0} \frac{\cancel{1} - \cancel{a^3} - 3a^2 h - 3ah^2 - h^3 - \cancel{1} + \cancel{a^3}}{h}
&& \text{Expand and combine like terms}\\
\\
\qquad g'(a) =& \lim \limits_{h \to 0} \frac{-3a^2 h - 3ah^2 - h^3}{h}
&& \text{Factor the numerator}\\
\\
\qquad g'(a) =& \lim \limits_{h \to 0} \frac{\cancel{h} (-3a^2 - 3ah - h^2)}{\cancel{h}}
&& \text{Cancel out like terms}\\
\\
\qquad g'(a) =& \lim \limits_{h \to 0} (- 3a^2 - 3ah - h^2) = -3a^2 - 3a(0) - (0)^2
&& \text{Evaluate the limit}\\
\\
\qquad g'(a) =& -3a^2
&& \text{Substitute value of $(a)$} \\
\\
\qquad g'(0) =& -3 (0)^2
&& \text{Simplify}\\
\\

\end{aligned}
\end{equation}
$


$\qquad \boxed{g'(0) = 0} \qquad$ Slope of the tangent line at $(0,1)$

Using Point Slope Form wehre the tangent line $y = g(x)$ at $(a, g(a))$


$
\begin{equation}
\begin{aligned}
y - g(a) =& g'(a)(x - a)
&& \\
\\
y - 1 =& 0 (x - 0)
&& \text{Substitute value of $a, g(a)$ and $g'(a)$}\\
\\
y - 1 =& 0
&& \text{Simplify}

\end{aligned}
\end{equation}
$


$\qquad \quad \boxed{ y = 1} \qquad $ Equation of the tangent line at $(0,1)$

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