Question

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

Answer 1

`y=-(2sqrt(10))/11 x+(15sqrt(10))/11`

Explanation:

We begin by finding the derivative of the function `f(x)` using the chain rule

`f'(x)=1/2*1/sqrt(x^3-x+4)*(3x^2-1)`

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

`f'(2)=1/2*1/sqrt(8-2+4)*(3*4-1)=11/(2sqrt(10))`

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

`m_p = -1//f'(2) = -(2sqrt(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(2)=sqrt(8-2+4)=sqrt(10)`

Substituting this into the equation of the line we obtain

`sqrt(10)=-(2sqrt(10)/11)*2+b`

Solving for `b` we get

`b=(15sqrt(10))/11`

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

`y=-(2sqrt(10))/11 x+(15sqrt(10))/11`