Question

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

Answer 1

`-1/4`

Explanation:

We know that a normal line is orthogonal to a tangent line, so we should find the slope of the tangent line first, then find the slope orthogonal to it.

The slope of the tangent line is just the derivative:
`f'(x) = 2x e^(x-1) + x^2 e^(x-1) + 2`
`f'(1) = 2e^0 + 1e^0 + 2 = 4`

Two orthogonal lines have
`m_1m_2 = -1`

so
`m_(norm) * 4 = -1 implies m_(norm) = -1/4`