Question

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

Answer 1

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 `(a, f(a))` can be defined in several ways (or using several notations). Two of the more common are:

`lim_(xrarra) (f(x)-f(a))/(x-a)` `" "` OR `" "` `lim_(hrarr0) (f(a+h)-f(a))/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(x) = 3x^2` and `a=1`

We'll find:

`lim_(xrarra) (f(x)-f(a))/(x-a)`
(Note that substitution gets us the indeterminate form `0/0`. We have some work to do.)

`lim_(xrarr1) (f(x)-f(1))/(x-1) = lim_(xrarr1) (3x^2-3)/(x-1)`

`= lim_(xrarr1) (3(x^2-1))/(x-1)`

`= lim_(xrarr1) (3(x+1)(x-1))/(x-1)`

The expression whose limit we want is equal to `3(x+1)` 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_(xrarr1) (3(x+1)(x-1))/(x-1) = lim_(xrarr1) 3(x+1) = 6`

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

Short method

For `f(x) = 3x^2`, we get `f'(x) = 3*2 x^(2-1) = 6x`

The slope of the tangent at `1` is `f'(1) = 6(1) =6`