# How do you find the slope of the tangent line for f(x) = 3x^2 at (1,3)?

Aug 29, 2015

The slope is $6$.

#### Explanation:

I will assume that you have not yet been taught the rules (shortcuts) for finding derivatives. So, we will use a definition.

The slope of the line tangent to the graph of the function $f$ at the point $\left(a , f \left(a\right)\right)$ can be defined in several ways (or using several notations). Two of the more common are:

${\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$ $\text{ }$ OR $\text{ }$ ${\lim}_{h \rightarrow 0} \frac{f \left(a + h\right) - f \left(a\right)}{h}$

(Each author,teacher,presenter needs to choose one definition as the 'official' definition. Many will immediately mention other possibilities as 'equivalents'.)

For this question we have $f \left(x\right) = 3 {x}^{2}$ and $a = 1$

We'll find:

${\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$
(Note that substitution gets us the indeterminate form $\frac{0}{0}$. We have some work to do.)

${\lim}_{x \rightarrow 1} \frac{f \left(x\right) - f \left(1\right)}{x - 1} = {\lim}_{x \rightarrow 1} \frac{3 {x}^{2} - 3}{x - 1}$

$= {\lim}_{x \rightarrow 1} \frac{3 \left({x}^{2} - 1\right)}{x - 1}$

$= {\lim}_{x \rightarrow 1} \frac{3 \left(x + 1\right) \left(x - 1\right)}{x - 1}$

The expression whose limit we want is equal to $3 \left(x + 1\right)$ for all $x$ other than $x = 1$. The limit doesn't care what happens when $x$ is equal to $1$, it wants to onow what happens when $x$ is close to $1$, so we get:

${\lim}_{x \rightarrow 1} \frac{3 \left(x + 1\right) \left(x - 1\right)}{x - 1} = {\lim}_{x \rightarrow 1} 3 \left(x + 1\right) = 6$

The slope of the tangent we were asked about is $6$.

Short method

For $f \left(x\right) = 3 {x}^{2}$, we get $f ' \left(x\right) = 3 \cdot 2 {x}^{2 - 1} = 6 x$

The slope of the tangent at $1$ is $f ' \left(1\right) = 6 \left(1\right) = 6$