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 $y = x + 4$