# How do I find the equation for a tangent line without derivatives?

Aug 4, 2018

You could use infinitesimals...

#### Explanation:

The slope of the tangent line is the instantaneous slope of the curve. So if we increase the value of the argument of a function by an infinitesimal amount, then the resulting change in the value of the function, divided by the infinitesimal will give the slope (modulo taking the standard part by discarding any remaining infinitesimals).

For example, suppose we want to find the tangent to $f \left(x\right)$ at $x = 2$, where:

$f \left(x\right) = {x}^{3} - 3 {x}^{2} + x + 5$

Let $\epsilon > 0$ be an infinitesimal value. Then:

$\frac{f \left(2 + \epsilon\right) - f \left(2\right)}{\epsilon}$

$= \frac{\left({\left(2 + \epsilon\right)}^{3} - 3 {\left(2 + \epsilon\right)}^{2} + \left(2 + \epsilon\right) + 5\right) - \left({\left(2\right)}^{3} - 3 {\left(2\right)}^{2} + \left(2\right) + 5\right)}{\epsilon}$

$= \frac{\left(\left(8 + 12 \epsilon + 6 {\epsilon}^{2} + {\epsilon}^{3}\right) - 3 \left(4 + 4 \epsilon + {\epsilon}^{2}\right) + \left(2 + \epsilon\right) + 5\right) - \left(8 - 12 + 2 + 5\right)}{\epsilon}$

$= \frac{\left(12 \epsilon + 6 {\epsilon}^{2} + {\epsilon}^{3}\right) - \left(12 \epsilon + 3 {\epsilon}^{2}\right) + \epsilon}{\epsilon}$

$= \frac{\epsilon + 3 {\epsilon}^{2} + {\epsilon}^{3}}{\epsilon}$

$= 1 + 3 \epsilon + {\epsilon}^{2}$

of which the standard (i.e. finite) part is $1$ (discarding the $3 \epsilon + {\epsilon}^{2}$).

So the slope of the tangent is $1$ and the tangent point is:

$\left(2 , f \left(2\right)\right) = \left(2 , 3\right)$

So the equation of the tangent may be written:

$\left(y - 3\right) = 1 \left(x - 2\right)$

or more simply:

$y = x + 1$

graph{ (y-(x^3-3x^2+x+5))(y-x-1) = 0 [-3.355, 6.645, 1.38, 6.38]}