# What is the equation of the normal line of f(x)=sqrt(x^3-x+4) at x=2?

Mar 9, 2016

$y = - \frac{2 \sqrt{10}}{11} x + \frac{15 \sqrt{10}}{11}$

#### Explanation:

We begin by finding the derivative of the function $f \left(x\right)$ using the chain rule

$f ' \left(x\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{{x}^{3} - x + 4}} \cdot \left(3 {x}^{2} - 1\right)$

The slope of the tangent line at $x = 2$ is then

$f ' \left(2\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{8 - 2 + 4}} \cdot \left(3 \cdot 4 - 1\right) = \frac{11}{2 \sqrt{10}}$

The slope of the normal line is therefore the negative reciprocal of this slope:

${m}_{p} = - 1 / f ' \left(2\right) = - \frac{2 \sqrt{10}}{11}$

To get the equation of the line, we need to find a point to substitute into the equation to get the y-intercept - we need the value of the function at $x = 2$

$f \left(2\right) = \sqrt{8 - 2 + 4} = \sqrt{10}$

Substituting this into the equation of the line we obtain

$\sqrt{10} = - \left(2 \frac{\sqrt{10}}{11}\right) \cdot 2 + b$

Solving for $b$ we get

$b = \frac{15 \sqrt{10}}{11}$

Therefore the equation of the normal line to our function at $x = 2$ is

$y = - \frac{2 \sqrt{10}}{11} x + \frac{15 \sqrt{10}}{11}$