How do you find the linearization of #f(x)=x^3# at the point x=2?

Question

Emilia Aguilar · Accepted Answer

The linearization is the equation of the tangent line. (Often presented in a different form.)

#f(x) = x^3#. At #x = 2#, we have #y = 8#. #f'(x) = 3x^2#, so at #x=2#, we have #f'(2) = 12# The linearization is #L(x) = 8+12(x-2)# The equation of the tangent line The line tangent to the graph at the point where #x=2# contains tha point #(2,8) and has slope #m=12#. In point-slope form the equation is #(y-8) = 12(x-2)# When thinking of it as a linear approximation, we write #y=8+12(x-2)# More explanation The linearization uses #y=8# as a starting point and adds the change in #y# along the tangent line for a particular change in #x#. The change in #y# along the tangent line is called the differential of #y# and is denoted #dy#. On a line of slope #m#, the change in #y# along the line is the slope times the change in #x#, in other words, the change in #y# along the line #= m Deltax# For the differential, we change the notation to #dx# and write: #dy=mdx# where #m = f(x)# at some chosen #x=a#. #dy = f'(a)Deltax = f'(a)(x-a)# In this question the chosen value of #x# was #a = 2#, so #dy = f'(a)(x-a) = 12(x-2)#. The approximation begins at #y = 8# and adds #dy# #L(x) = f(a) + f'(a)(x-a) = 8 + 12(x-2)#

Charles Anctil · Answer

undefined To find the linearization of $ f(x) = x^3 $ at the point $ x = 2 $, follow these steps:

1. Find the value of the function $ f(x) $ at the given point $ x = 2 $: $ f(2) = 2^3 = 8 $.

2. Find the derivative of the function $ f'(x) = 3x^2 $.

3. Evaluate the derivative at the point $ x = 2 $: $ f'(2) = 3(2)^2 = 12 $.

4. Write the equation of the tangent line using the point-slope form: 
   $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point $ (2, 8) $ and $ m $ is the slope $ f'(2) $.

5. Substitute the values into the equation: 
   $ y - 8 = 12(x - 2) $.

6. Simplify the equation: 
   $ y - 8 = 12x - 24 $.

7. Rearrange the equation to isolate $ y $: 
   $ y = 12x - 24 + 8 $.

8. Simplify further: 
   $ y = 12x - 16 $.

So, the linearization of $ f(x) = x^3 $ at the point $ x = 2 $ is $ L(x) = 12x - 16 $.