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

Question

Luke Alvarez · Accepted Answer

The slope is #6#.

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#

Luke Alvarez · Answer

undefined To find the slope of the tangent line for $ f(x) = 3x^2 $ at $ (1,3) $, you can use the derivative of the function. The derivative of $ f(x) $ with respect to $ x $ is $ f'(x) = 6x $. Evaluate $ f'(1) $ to find the slope of the tangent line at $ x = 1 $. $ f'(1) = 6 $. Therefore, the slope of the tangent line for $ f(x) = 3x^2 $ at $ (1,3) $ is $ 6 $.

Elijah Abram · Answer

undefined To find the slope of the tangent line for the function $ f(x) = 3x^2 $ at the point (1,3), you need to find the derivative of the function $ f(x) $ with respect to $ x $, and then evaluate the derivative at $ x = 1 $.

The derivative of $ f(x) = 3x^2 $ is $ f'(x) = 6x $.

Evaluate the derivative at $ x = 1 $ to find the slope of the tangent line at the point (1,3):

$$ f'(1) = 6(1) = 6 $$

So, the slope of the tangent line to the function $ f(x) = 3x^2 $ at the point (1,3) is 6.