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#
