Question

What is the equation of the line that is normal to `f(x)= xe^(2x-2)` at `x= 1`?

Answer 1

I found: `y=-1/3x+4/3`

Explanation:

We can first find the coordinates of the point of intersection: knowing that `x=1` we substitute into our function and it gives us:
`y=f(1)=1*e^(2*1-2)=1*e^0=1`
So we get `(1,1)`.

Next we try to evaluate the GRADIENT of the line perpendicular (normal).
We evaluate the derivative of the function:
`f'(x)=1*e^(2x-2)+2xe^(2x-x)=e^(2x-x)[2x+1]`
Evaluate the derivative at `x=1`
`f'(1)=1[2+1]=3`
This is the slope `m_T` of the TANGENT to your curve at your point. To finde the slope of the normal `m_N` we need:
`m_N=-1/m_T=-1/3`

Finally we use the general equation of a line through a point and having slope equal to `m_T`:
`y-y_0=m_T(x-x_0)`
`y-1=-1/3(x-1)`
`y=1-1/3x+1/3`
`y=-1/3x+4/3`

Graphically: