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
