Question

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

Answer 1

`-1/2`

Explanation:

In this case, the slope of the tangent line will be given by `f'(2).`

The normal line is the line perpendicular to the tangent line; therefore, its slope will be the negative reciprocal of the slope to the tangent line, or,

`m_n=-1/(f'(2))`

Let's first differentiate the function:

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

Determine the slope of the tangent:

`f'(2)=e^(2-2)+1=e^0+1=2`

The slope of the normal line is then

`-1/(f'(2))=-1/2`