Question

What is the equation of the normal line of `f(x)=e^-x+x^3` at `x=-5`?

Answer 1

`y = -x/(e^5-75) +e^5-125-5/(e^5-75)`

Explanation:

`f(x)=e^-x+x^3`

When `x=-5 => f(-5)=e^5-125`

Differentiating wrt `x`;
`f(x)=-e^-x+3x^2`
so When `x=-5 => f'(-5)=-e^5+75 =75-e^5`

This is the slope of the tangent when `x=-5`, as the normal is perpendicular to the tangent, their product is `-1`

So, the slope of the normal is `-1/(75-e^5)=1/(e^5-75)`

Using `y-y_1=m(x-x_1)`, the normal equation is;

`y-(e^5-125) = 1/(e^5-75)(x-(-5))`
`y-e^5+125 = 1/(e^5-75)(x+5)`
`y-e^5+125 = x/(e^5-75)+5/(e^5-75)`
`y = x/(e^5-75) +e^5-125+5/(e^5-75)`