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

1 Answer
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 (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