Question

What is the slope of the line normal to the tangent line of `f(x) = xe^(x-1)+x^3` at `x= 2`?

Answer 1

Slope of normal line to the tangent would be `-1/ (3e+12`

Explanation:

Slope of tangent line is `f' (x)= xe^(x-1) +e^(x-1) +3x^2`

at x=2, slope would be f'(2)= 3e +12

Slope of normal line to the tangent would be `-1/ (3e+12`