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

1 Answer
Mar 9, 2016

#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#