Question

How do you find the slope of the tangent line to the graph `f(x)=6x` when `x=3`?

Answer 1

I will assume that you are beginning your study of calculus and you want to do this using the definition.

You find `lim_(x rarr a) (6x-6a)/(x-a)`

By definition, the slope of the line tangent to the graph of `y=f(x)` at `x=a` is the limit, as `x` approaches `a`, of the slopes of the secant lines through the points (on the graph) `(a, f(a))` and `(x, f(x))`.

(This is the same as saying, the slope of the tangent line is the limit of `(Delta y)/(Delta x)` as `Delta xrarr0`.)

Formally, the slope of the tangent line to the graph of `f(x)=6x` at the point where `x=3` is `lim_(Delta x rarr 0) (Delta y)/(Delta x)`

This can be written `lim_(x rarr a) (f(x)-f(a))/(x-a)` which, for the

question asked here is `=lim_(x rarr 3) (6x-6(3))/(x-3)=lim_(x rarr 3) (6x-18)/(x-3)`

Observe that we cannot evaluate this limit by substitution, because when `x=a`, the denominator is `0`. So we write:

`lim_(x rarr 3) (6x-18)/(x-3)=lim_(x rarr 3) (6(x-3))/(x-3)=lim_(x rarr 3) 6 = 6`.

(Later, you'll learn shortcuts, but it is important to understand the reasoning that underlies the shortcuts.)

Alternate notation Sometimes the algebra is simpler if, instead of naming the changing value "`x`", we name the difference (`x-a`). Some like to call this difference `Delta x` another common notation is to call it `h`.

In this notation the reasoning above looks like this:

The slope of the tangent line at `x=3` is

`lim_(h rarr 0)(f(3+h)-f(3))/h=lim_(h rarr 0)(6(3+h)-6(3))/h`

`=lim_(h rarr 0)(18+6h-18)/h=lim_(h rarr 0)(6h)/h=lim_(h rarr 0) 6 = 6`