What is the equation of the line normal to f(x)=ln(xe^(3x))f(x)=ln(xe3x) at x=1x=1?

1 Answer
iceman
Apr 22, 2018

y=(-1/4)x+13/4y=(−14)x+134

Explanation:

f(x) = ln(xe^(3x))f(x)=ln(xe3x)
f(1)= ln(1*e^3)=3ln(e)=3f(1)=ln(1⋅e3)=3ln(e)=3
f(x)=ln(x)+3x*ln(e)=ln(x)+3xf(x)=ln(x)+3x⋅ln(e)=ln(x)+3x
f'(x)=1/x+3
f'(1)=4 => slope of the tangent at x = 1:
Slope of the normal at x = 1 is: -1/4:
m=-1/4, (x_1,y_1)=(1,3)
y-y_1=m(x-x_1)
y-3=-1/4(x-1)
y=(-1/4)x+13/4

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