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

Question

Emilia Aguilar · Accepted Answer

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#

Cameron Alexander · Answer

undefined To find the slope of the tangent line to the graph of f(x) = 6x when x = 3, we can use the derivative of the function. The derivative of f(x) = 6x is f'(x) = 6. Therefore, the slope of the tangent line at x = 3 is 6.