Question

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

Answer 1

`:.x=-1`

Explanation:

The slope `m_n=-1/m_(x_0)`

`m_(x_0)=f'(x_0)`

and the generic line equation is:

`(y-y_0)=m_n(x-x_0)`

with `y_0=f(x_0)`

• We have to find `f'(x)` using the Chain Rule

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

`f'(x_0)=f'(-1)=2(-1+1)=0`

`:.x_0=-1` it's a critical point of `f(x)`

`m_n=-1/m_(x_0)=-1/0 =>cancel EE`

then in `x_0` the normal line has no slope and it's

`x=x_0`

`:.x=-1`

Alternatively:

`f(x)=(x+1)^2`

It's a parabola with two cincident real roots.

The roots are `x=-1=x_0`

Then the vertex of the parabola it's `V(-1,0)`. The vertex it's a local min (because if we rewrite `f(x)` as `y=ax^2+bx+c` we have `a>0` ), where `f'(x)=0`. The tangent line it's `y=0` therefore the normal it's `x=-1`