What is the equation of the normal line of #f(x)=(x-2)^(3/2)-x^3# at #x=2#?

Question

Emma Aldridge · Accepted Answer

The Normal Line is

#y=x/12-49/6# or #x-12y=98# the given: #f(x) =(x-2)^(3/2)-x^3# and #x=2# #f(2) =(2-2)^(3/2)-(2)^3# #f(2)= -8# the point on the curve : #(2, -8)# compute slope #m# then use #-1/m# for the normal line #f' (x) = (3/2)(x-2)^(1/2)-3x^2# #f' (2) = (3/2)(2-2)^(1/2)-3(2)^2# #f' (2) = -12# The slope to be used to find the normal line is #-1/m# #-1/m=-1/-12=1/12# Determine now the Normal Line using #y-y_1=(1/12)(x-x_1)# #y-(-8)=(1/12)(x-2)# after simplification the final answer: #x-12y=98# or #y=x/12-49/6#

Ezra Adams · Answer

undefined The equation of the normal line of f(x)=(x-2)^(3/2)-x^3 at x=2 is y = -4x + 8.

Ezra Adams · Answer

undefined The equation of the normal line to the function $ f(x) = (x - 2)^{\frac{3}{2}} - x^3 $ at $ x = 2 $ can be found by first determining the slope of the tangent line at $ x = 2 $ and then using the negative reciprocal of this slope to find the slope of the normal line. Finally, the equation of the normal line can be written in point-slope form using the point of tangency $ (2, f(2)) $.

First, find $ f'(x) $, the derivative of $ f(x) $, and evaluate it at $ x = 2 $ to find the slope of the tangent line. Then, take the negative reciprocal of this slope to find the slope of the normal line. Finally, use the point-slope form of a line to write the equation of the normal line.

$ f'(x) = \frac{3}{2}(x - 2)^{\frac{1}{2}} - 3x^2 $

Evaluate $ f'(2) $ to find the slope of the tangent line:

$ f'(2) = \frac{3}{2}(2 - 2)^{\frac{1}{2}} - 3(2)^2 $
$ f'(2) = -12 $

The slope of the normal line is the negative reciprocal of the slope of the tangent line:

$ m_{\text{normal}} = \frac{-1}{-12} = \frac{1}{12} $

Using the point-slope form of a line with the point $ (2, f(2)) $ and the slope $ \frac{1}{12} $, the equation of the normal line is:

$ y - f(2) = \frac{1}{12}(x - 2) $

Substitute $ x = 2 $ into $ f(x) $ to find $ f(2) $:

$ f(2) = (2 - 2)^{\frac{3}{2}} - 2^3 $
$ f(2) = 0 - 8 = -8 $

So the equation of the normal line is:

$ y + 8 = \frac{1}{12}(x - 2) $