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

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Answer 1 · 2016-02-01T18:47:38

Slope of normal line to the tangent would be $- \frac{1}{3 e + 12}$ $-1/ (3e+12$

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 $-1/ (3e+12$

What is the slope of the line normal to the tangent line of f(x) = xe^(x-1)+x^3 f(x)=xex−1+x3f(x) = xe^(x-1)+x^3 at x= 2 x=2 x= 2 ?