What is the equation of the normal line of #f(x)= x^2/(x^3− 5x + 1)# at #x = 5#?

Question

1291 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-26T12:53:24

Equation of normal line is # y= 13.79x - 68.7#

#f(x)=x^2/(x^3-5x+1) :. f(5)=5^2/(5^3-5*5+1) ~~0.25(2dp) #

Point is at #(x_1,y_1) =(5,0.25)# . Slope is #m= f'(x)#

#f'(x) = (2x*(x^3-5x+1)- x^2(3x^2-5))/(x^3-5x+1)^2#

#f'(x) =m= (25*(5^3-5*5+1)- 5^2(3*5^2-5))/(5^3-5*5+1)^2# or

#m ~~-0.07 #. Slope of normal is #m_1= -1/-0.07~~13.79(2dp)#

Equation of normal line is #(y-y_1)=m_1(x-x_1)# or

#(y-0.25)=13.79(x-5) or y= 13.79x - 68.7# [Ans]