What is the equation of the line that is normal to #f(x)= x^3e^sqrtx # at # x= 2 #?

Question

What is the equation of the line that is normal to #f(x)= x^3e^sqrtx   # at #  x= 2 #?

Cameron Alexander · Accepted Answer

#y-8e^sqrt2=frac{-1}{[2e^(sqrt2)(2^(1/2)+6)]}(x-2)#
#y=frac{-x}{2e^sqrt2(sqrt2+6)}+frac{1}{e^sqrt2(sqrt2+6)}+8e^sqrt2# First, find the equation of the tangent line to #f(x)# at x=2 in point slope form, then write the normal line equation the same, except with an opposite reciprocal slope. To find slope of tangent line: Find #f'(x)#: #f(x)=x^3e^(sqrtx)# #f'(x)=(x^3)(1/2x^(-1/2)e^(sqrtx))+(3x^2)(e^(sqrtx))# #f'(x)=[x^3e^(sqrtx)]/[2sqrtx]+3x^2e^(sqrtx)# #f'(x)=(x^2e^sqrtx)(1/2x^(1/2)+3)# #f'(x)=(1/2x^2e^(sqrtx))(x^(1/2)+6)# #f'(2)=(1/2(2)^2e^(sqrt(2)))((2)^(1/2)+6)# #f'(2)=[2e^(sqrt2)(2^(1/2)+6)]# Therefore, slope of normal line: #frac{-1}{f'(2)}=frac{-1}{[2e^(sqrt2)(2^(1/2)+6)]}# Point: #f(2)=2^3e^sqrt2# #=8e^sqrt2# #(2,8e^sqrt2)# Equation of normal line: #y-y_1=m(x-x_1)# #y-8e^sqrt2=frac{-1}{[2e^(sqrt2)(2^(1/2)+6)]}(x-2)# #y=frac{-1}{2e^sqrt2(2^(1/2)+6)}(x-2)+8e^sqrt2# #y=frac{-x}{2e^sqrt2(sqrt2+6)}+frac{1}{e^sqrt2(sqrt2+6)}+8e^sqrt2#

Paisley Johnson · Answer

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