Question

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

Answer 1

`-e/(3+12e)`

Explanation:

Before anything else, find the function's derivative. To do this, we will need the product rule for the `xe^(x-3)` term. For `x^3`, just use the power rule.

`f'(x)=e^(x-3)d/dx(x)+xd/dx(e^(x-3))+3x^2`

The simpler derivative here is `d/dx(x)=1`. The more difficult is `d/dx(e^(x-3))`, for which we will need to use chain rule.

Since `d/dx(e^x)=e^x`, we can say that `d/dx(e^(g(x)))=e^(g(x))*g'(x)`.

Thus, `d/dx(e^(x-3))=e^(x-3)d/dx(x-3)=e^(x-3)(1)=e^(x-3)`.

Plugging this back into our `f'(x)` equation, we see that

`f'(x)=e^(x-3)(1)+x(e^(x-3))+3x^2`

`f'(x)=e^(x-3)(x+1)+3x^2`

Now, we must find the slope of the tangent line, which is equal to the value of the derivative at `x=2`. From here, we will be able to find the slope of the normal line, since the two are perpendicular.

`f'(2)=e^(2-3)(2+1)+3(2^2)=e^-1(3)+3(4)=3/e+12`

This, however, is the slope of the tangent line. Since perpendicular lines have opposite reciprocal slopes, the slope of the normal line is

`-1/(3/e+12)=-1/((3+12e)/e)=-e/(3+12e)`