What is the slope of the line normal to the tangent line of #f(x) = 2x-4sqrt(x-1) # at # x= 2 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = 2x-4sqrt(x-1)  # at # x= 2  #?

Charles Anctil · Accepted Answer

The slope will be undefined.

Start by finding the y-coordinate of the point of tangency. #f(2) = 2(2) - 4sqrt(2 - 1)# #f(2) = 4 - 4# #f(2) = 0# Find the derivative of #f(x)#. #f'(x) = 2 - 4/(2sqrt(x - 1))# #f'(x) = 2 - 2/sqrt(x - 1)# Now find the slope of the tangent. #f'(2) = 2 - 2/sqrt(2 - 1) = 2 - 2/1 = 0# The normal line is perpendicular to the tangent line. The slope of #0# of the tangent line means the line will be #y = a#, where #a# is a constant. Then the line perpendicular to this will be of the form #x = b#, where the slope is undefined. Hopefully this helps!

Axel Alvarez · Answer

undefined To find the slope of the line normal to the tangent line of the function f(x) = 2x - 4√(x - 1) at x = 2, we first need to find the slope of the tangent line. Then, we can find the negative reciprocal of this slope to get the slope of the normal line.

1. Find the derivative of the function f(x) with respect to x.
2. Evaluate the derivative at x = 2 to find the slope of the tangent line.
3. Take the negative reciprocal of this slope to find the slope of the normal line.

Let's find the derivative of f(x):

f'(x) = d(2x - 4√(x - 1))/dx

To find the derivative of 2x, we get 2.

To find the derivative of -4√(x - 1), we apply the chain rule:

d/dx(√u) = (1/2√u) * du/dx

Here, u = (x - 1), so du/dx = 1.

Substitute u = (x - 1) and du/dx = 1:

d/dx(-4√(x - 1)) = -4 * (1/2√(x - 1)) * 1 = -2/√(x - 1)

So, f'(x) = 2 - 2/√(x - 1)

Now, evaluate f'(x) at x = 2:

f'(2) = 2 - 2/√(2 - 1) = 2 - 2/√1 = 2 - 2 = 0

The slope of the tangent line at x = 2 is 0.

The slope of the normal line is the negative reciprocal of the slope of the tangent line, so it is undefined.