How do you find the equation of the tangent line to #f(x) = (x-1)^3# at the point where #x=2#?

Question

Cameron Alexander · Accepted Answer

First of all, let's evaluate the function in #x=2#: we have #(2-1)^3=1^3=1#. So, we are looking for a line passing through the point #(2,1)#. The general equation of a line passing through a point #(x_0,y_0)# is
#y-y_0=m(x-x_0)#. So, in our, case, we have the equation
#y-1=m(x-2)#. Now we need to find the slope #m#, which is by definition the value of the derivative in #x=2#. Using the power rule, which states that #D f^n(x)= n f^{n-1}(x) f'(x)#, we have that #f'(x)=3(x-1)^2#. This means that #f'(2)=3#. We thus have everything we need to answer: the line is
#y-1=3(x-2)#, which we can bring into the standard form writing
#y=3x-5# Here's a link where you can see the graph.

Elijah Abram · Answer

undefined To find the equation of the tangent line to f(x) = (x-1)^3 at the point where x=2, we need to find the slope of the tangent line and the coordinates of the point of tangency.

First, we find the derivative of f(x) using the power rule: f'(x) = 3(x-1)^2.

Next, we substitute x=2 into the derivative to find the slope of the tangent line at x=2: f'(2) = 3(2-1)^2 = 3.

Now, we find the y-coordinate of the point of tangency by substituting x=2 into the original function: f(2) = (2-1)^3 = 1.

Therefore, the point of tangency is (2, 1) and the slope of the tangent line is 3.

Using the point-slope form of a linear equation, we can write the equation of the tangent line as y - 1 = 3(x - 2). Simplifying, we get y = 3x - 5.

Thus, the equation of the tangent line to f(x) = (x-1)^3 at the point where x=2 is y = 3x - 5.