How do you use the limit definition to find the slope of the tangent line to the graph #f(x) = (x-1)^3# at x=2?

1 Answer
Jan 16, 2016

The slope is #3#. Please see the explanation section, below.

Explanation:

Although we could use the defintion in the form #lim_(hrarr0)(f(x+h)-f(x))/h#,

I think that the algebra is simpler if we use the equivalent

The slope of the tangent line to the graph of #f(x)# at the point where #x=2# is

#lim_(xrarr2)(f(x)-f(2))/(x-2)#.

In this case, we get:

#lim_(xrarr2)((x-1)^3-(2-1)^3)/(x-2)#.

Obviously (as expected), the initial form of this limit is indeterminate, so we'll need to reduce the fraction and try again.

Expanding #(x-1)^3# either by multiplying or by the binomial expansion, and then simplifying, we get

#lim_(xrarr2)((x^3-3x^2+3x-1)-(1))/(x-2) = lim_(xrarr2)(x^3-3x^2+3x-2)/(x-2)#

Because #2# is a zero of the polynomial in the numerator, we know that #x-2# is a factor. By division, or by trial and error or grouping, we get

#x^3-3x^2+3x-2 = x^3-2x^2-x^2+2x+x-2#

# = (x-2)(x^2-x+1)#.

So our limit becomes:

#lim_(xrarr2)((x-1)^3-(2-1)^3)/(x-2) = lim_(xrarr2)((x^3-3x^2+3x-1)-(1))/(x-2)#

# = lim_(xrarr2)(x^3-3x^2+3x-2)/(x-2)#

# = lim_(xrarr2)(cancel((x-2))(x^2-x+1))/cancel((x-2))#

# = 4-2+1#

# =3#