Question

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

Answer 1

The equation is `y = 2x + 3`.

Explanation:

We start by finding the y-coordinate at the given x-point.

`f(-1) = sqrt(1/(-1 + 2)) = sqrt(1) = 1`

We can start by simplifying the function.

`f(x) = 1/sqrt(x + 2)`

We now use the chain rule to differentiate `sqrt(x + 2)` before using the quotient rule to differentiate `1/sqrt(x + 2)`.

Let `y = u^(-1/2)` and `u = x + 2`. Then `dy/(du) = -1/(2u^(3/2))` and `(du)/dx = 1`.

`dy/dx= dy/(du) xx (du)/dx`

`dy/dx = 1 xx -1/(2u^(3/2))`

`dy/dx = -1/(2(x + 2)^(3/2))`

The slope of the tangent is given by evaluating `x= a` inside the derivative.

`m_"tangent" = -1/(2(-1 + 2)^(3/2))`

`m_"tangent" =- 1/(2(1)^(1/2))`

`m_"tangent" =- 1/2`

The normal line is perpendicular to the tangent, that's to say its slope is the negative reciprocal of that of the tangent.

`m_"normal" = -1/(m_"tangent")`

`m_"normal" = -1/(-1/2)`

`m_"normal" = 2`

The equation of the normal line can be found using a point `(-1, 1)` and the slope `(2)`.

`y - y_1 = m(x- x_1)`

`y - 1= 2(x - (-1))`

`y - 1 = 2(x + 1)`

`y - 1= 2x + 2`

`y = 2x + 3`

Hopefully this helps!

Answer 2

`2x-y+3=0`. The inserted Socratic graph depicts all the features in the explanation.

Explanation:

To make f real, `x > -2`.

At `x =-1, y = 1`. And the foot of the normal is `P(-1, 1)`.

`f'=((x+2)^(-1/2))' =-1/2(1/((x+2)sqrt(x+2)))`. So, the slope of the normal is

`-1/(f')=2(x+2)sqrt(x+2)=2`, at `P(-1, 1)`.

Its equation is

`y-1=2(x-(-1))`, giving

`2x-y+3=0`..

graph{(ysqrt(x+2)-1)(2x-y+3)=0x^2 [-10, 10, -5, 5]}