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

Jan 2, 2017

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

#### Explanation:

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

$f \left(- 1\right) = \sqrt{\frac{1}{- 1 + 2}} = \sqrt{1} = 1$

We can start by simplifying the function.

$f \left(x\right) = \frac{1}{\sqrt{x + 2}}$

We now use the chain rule to differentiate $\sqrt{x + 2}$ before using the quotient rule to differentiate $\frac{1}{\sqrt{x + 2}}$.

Let $y = {u}^{- \frac{1}{2}}$ and $u = x + 2$. Then $\frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{2 {u}^{\frac{3}{2}}}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 1$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 \times - \frac{1}{2 {u}^{\frac{3}{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 {\left(x + 2\right)}^{\frac{3}{2}}}$

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

${m}_{\text{tangent}} = - \frac{1}{2 {\left(- 1 + 2\right)}^{\frac{3}{2}}}$

${m}_{\text{tangent}} = - \frac{1}{2 {\left(1\right)}^{\frac{1}{2}}}$

${m}_{\text{tangent}} = - \frac{1}{2}$

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")

${m}_{\text{normal}} = - \frac{1}{- \frac{1}{2}}$

${m}_{\text{normal}} = 2$

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

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 1 = 2 \left(x - \left(- 1\right)\right)$

$y - 1 = 2 \left(x + 1\right)$

$y - 1 = 2 x + 2$

$y = 2 x + 3$

Hopefully this helps!

Jan 2, 2017

$2 x - y + 3 = 0$. 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 $P \left(- 1 , 1\right)$.

$f ' = \left({\left(x + 2\right)}^{- \frac{1}{2}}\right) ' = - \frac{1}{2} \left(\frac{1}{\left(x + 2\right) \sqrt{x + 2}}\right)$. So, the slope of the normal is

$- \frac{1}{f '} = 2 \left(x + 2\right) \sqrt{x + 2} = 2$, at $P \left(- 1 , 1\right)$.

Its equation is

$y - 1 = 2 \left(x - \left(- 1\right)\right)$, giving

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

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