What is the equation of the line normal to #f(x)=4/(2x-1) # at #x=0#?

1 Answer
Jul 20, 2016

A normal line is a line that is perpendicular to the tangent. In other words, we must first find the equation of the tangent.

Explanation:

Step 1: Determine which point the function and the tangent pass through

#f(0) = 4/(2 xx 0 - 1)#

#f(0) = 4/(-1)#

#f(0) = -4#

#:.# The function passes through #(0, -4)#

Step 2: Differentiate the function

Let #f(x) = (g(x))/(h(x))#.

Then, #g(x) = 4# and #h(x) = 2x - 1#. The derivative is given by #dy/dx = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#

The derivative of #g(x) = 4# is #g'(x) = 0#. The derivative of #h(x) = 2x - 1# is #2#.

We can now use the quotient rule, as shown above, to determine the derivative.

#dy/dx = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#

#dy/dx = (0 xx (2x - 1) - 4(2))/(2x - 1)^2#

#dy/dx = -8/(2x - 1)^2#

Step 3: Determine the slope of the tangent

The slope of the tangent is given by evaluating #f(a)# inside the derivative, where #x= a# is the given point.

Then, we can say:

#m_"tangent" = -8/(2 xx 0 - 1)^2 = (-8)/(1) = -8#

Step 4: Determine the slope of the normal line

As mentioned earlier, the normal line is perpendicular, but passes through the same point of tangency that does the tangent. A line perpendicular to another has a slope that is the negative reciprocal of the other. The negative reciprocal of #-8# is #1/8#. Thus, the slope of the normal line is #1/8#.

Step 5: Determine the equation of the normal line using point-slope form

We now know the slope of the normal line as well as the point of contact. This is enough for us to determine its equation using point-slope form.

#y - y_1 = m(x - x_1)#

#y - (-4) = 1/8(x - 0)#

#y + 4 = 1/8x - 0#

#y = 1/8x - 4#

In summary...

The line normal to #f(x) = 4/(2x - 1)# at the point #x = 0# has an equation of #y = 1/8x - 4#.

Practice exercises:

Determine the equation of the normal lines to the given relations at the indicated point.

a) #f(x) = (2x^2 + 6)/(3x^2 - 3)#, at #x = 3#

b) #f(x) = e^(x^2 - 4) + 2x^7#, at the point #x = -2#

c) #f(theta) = sin2theta#, at the point #theta = pi/4#

d) #f(x) = (x - 2)^2 + (y + 3)^2 =53# at the point #(9, -1)#

Hopefully this helps, and good luck!