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

Question

Paisley Johnson · Accepted Answer

#77y=-343x+996# First we find the y-coordinates; Substitute x=3 #3/(2-(3)^2)=-3/7# Then differentiate #x/(2-x^2)# #dy/dx#=#[x^2+2]/(x^2-2)^2# Substitute 3 to find the gradient of the tangent #dy/dx#=#[3^2+2]/[3^2-2]^2# =#11/49# General equation of normal: #(y-y_1)#=-1/(gradient of tangent)#(x-x_1)# Substituting the values #(y-(-3/7))=-1/(11/49)(x-3)# #y+3/7=-49/11x+147/11# #y=-49/11x+996/77# #77y=-343x+996#

Emily Ammerman · Answer

undefined To find the equation of the normal line of f(x) = x/(2-x^2) at x = 3, we need to determine the slope of the tangent line at x = 3 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) using the quotient rule:

f'(x) = [(2-x^2)(1) - (x)(-2x)] / (2-x^2)^2

Simplifying this expression, we get:

f'(x) = (2 - x^2 + 2x^2) / (2-x^2)^2
       = (2 + x^2) / (2-x^2)^2

Next, we substitute x = 3 into f'(x) to find the slope of the tangent line at x = 3:

f'(3) = (2 + 3^2) / (2-3^2)^2
      = (2 + 9) / (2-9)^2
      = 11 / (-7)^2
      = 11 / 49

Now, we take the negative reciprocal of this slope to find the slope of the normal line:

Slope of normal line = -1 / (11 / 49)
                    = -49 / 11

Finally, we use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope of the line. We substitute x = 3 and the slope of the normal line into this equation:

y - f(3) = (-49 / 11)(x - 3)

Simplifying further, we have:

y - (3/(2-3^2)) = (-49 / 11)(x - 3)

This is the equation of the normal line of f(x) = x/(2-x^2) at x = 3.