Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 6

Determine the equation of the tangent line to the curve $y = 2x^3 - 5x$ at the point ( -1, 3)

Using the definition (Slope of tangent line)


$
\begin{equation}
\begin{aligned}


\displaystyle m =& \lim \limits_{x \to a} \frac{f(x) - f(a)}{x - a}\\
\\
\text{We have } a =& -1 \text{ and } f(x) = 2x^3 - 5x, \text{ So the slope is }\\
\\
\displaystyle m =& \lim \limits_{x \to -1} \frac{f(x) - f(-1)}{x + 1}\\
\\
\displaystyle m =& \lim \limits_{x \to -1} \frac{2x^3 - 5x - [2 (-1)^3 - 5(-1)]}{x + 1}
&& \text{ Substitute value of $a$ and $x$}\\
\\
\displaystyle m =& \lim \limits_{x \to -1} \frac{(2x^3 - 5x - 3)}{x + 1}

\end{aligned}
\end{equation}
$



Using Synthetic Division to obtain the factor of the numerator










$
\begin{equation}
\begin{aligned}

\displaystyle m =& \lim\limits_{x \to -1} \frac{ (2x^2 - 2x - 3)\cancel{( x + 1)}}{\cancel{x + 1}}
&& \text{ Cancel out like terms}\\
\\
\displaystyle m =& \lim \limits_{x \to -1} (2x^2 - 2x - 3) = 2(-1)^2 - 2(-1)-3
&& \text{ Evaluate the limit }\\
\\
m =& 1

\end{aligned}
\end{equation}
$


Using point slope form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m ( x - x_1)\\
\\
y - 3 =& 1 ( x + 1 )
&& \text{ Substitute value of $x, y$ and $m$}\\
\\
y - 3 =& x + 1
&& \text{ Combine like terms}\\
\\
y =& x +4
\end{aligned}
\end{equation}
$



Therefore,
The equation of the tangent line at (-1,3) is $y = x + 4$