What is the equation of the line normal to # f(x)=-xln(2x^3-x)# at # x=1#?

Question

What is the equation of the line normal to #  f(x)=-xln(2x^3-x)# at #  x=1#?

Isaiah Allen · Accepted Answer

# y = 1/5x - 1/5 #

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is #-1#

so If # f(x )= -xln(2x^3-x) # then differentiating wrt #x# (using the product rule and chain rule) gives us:

# \ \ \ \ \ f'(x) = (-x)(1/(2x^3 - x)*(6x^2 - 1)) + (-1)(ln(2x^3 - x)) #
# :. f'(x) = -(6x^2-1)/(2x^2-1) -ln(2x^3-x) #

When #x = 1 => f(1) = -1ln(2-1) = 0 # (so #(1,0)# lies on the curve)
and # f'(1) = -(6-1)/(2-1) - ln(2-1) = -5#

So the tangent passes through #(1,0)# and has gradient #-5#, hence the normal has gradient #1/5# so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek is;

# y-0 = 1/5(x-1) #
# \ \ :. y = 1/5x - 1/5 #

We can confirm this solution is correct graphically:

Charles Anctil · Answer

undefined To find the equation of the line normal to the function f(x) = -xln(2x^3 - x) at x = 1, we need to determine the slope of the normal line and the point of tangency.

First, we find the derivative of f(x) using the product rule and chain rule:

f'(x) = -ln(2x^3 - x) - x * (1/(2x^3 - x)) * (6x^2 - 1)
       = -ln(2x^3 - x) - (6x^3 - x)/(2x^3 - x)
       = -ln(2x^3 - x) - (6x^3 - x)/(2x^3 - x)
       = -ln(2x^3 - x) - 6x^3 + x

Next, we substitute x = 1 into f'(x) to find the slope of the tangent line at x = 1:

f'(1) = -ln(2(1)^3 - 1) - 6(1)^3 + 1
      = -ln(2 - 1) - 6 + 1
      = -ln(1) - 5
      = -5

Since the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is 1/5.

Now, we have the slope of the normal line and the point of tangency (x = 1, f(1)). We can use the point-slope form of a line to find the equation of the normal line:

y - f(1) = (1/5)(x - 1)

Simplifying further:

y - (-1ln(2(1)^3 - 1)) = (1/5)(x - 1)
y + ln(1) = (1/5)(x - 1)
y = (1/5)(x - 1) - ln(1)
y = (1/5)(x - 1)

Therefore, the equation of the line normal to f(x) = -xln(2x^3 - x) at x = 1 is y = (1/5)(x - 1).