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

Question

William Abdalla · Accepted Answer

# y=2x-37/4#

We have # f(x) = (x-2)/(x-3)^2 #

Using the quotient rule we have:

# f'(x) = { (x-3)^2 d/dx(x-2) - (x-2)d/dx(x-3)^2 }/ ((x-3)^2)^2#
# :. f'(x) = { (x-3)^2 (1) - (x-2)2(x-3)(1) }/ ((x-3)^4)#
# :. f'(x) = { (x-3){ (x-3) - 2(x-2)} }/ ((x-3)^4)#
# :. f'(x) = { (x-3 - 2x+4) }/ ((x-3)^3)#
# :. f'(x) =(1-x)/ ((x-3)^3)#

When # x=5 => f'(5) = (1-5)/(5-3)^3 = -4/2^3 = -1/2 #

So the gradient of the tangent when #x=5# is #m_T=1/2#

The tangent and normal are perpendicular so the product of their gradients is #-1.#

So the gradient of the normal when #x=5# is #m_N=2#

Also #f(5)=(5-2)/(5-3)^2 = 3/2^2 = 3/4#

So the normal passes through (5,3/4) and has gradient #2#, so using #y=y_1=m(x-x_1)# the required equation is;

# y-3/4=2(x-5)#
# y-3/4=2x-10#
# y=2x-37/4#

Emilia Aguilar · Answer

undefined To find the equation of the normal line of f(x)=(x-2)/(x-3)^2 at x=5, we need to determine the slope of the tangent line at x=5 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can take the derivative of f(x) with respect to x.

The derivative of f(x) is given by f'(x) = [(x-3)^2 - 2(x-2)(2(x-3))]/(x-3)^4.

Evaluating f'(x) at x=5, we get f'(5) = [(5-3)^2 - 2(5-2)(2(5-3))]/(5-3)^4.

Simplifying this expression, we find f'(5) = 1/4.

Since the slope of the tangent line is 1/4, the slope of the normal line will be the negative reciprocal, which is -4.

Now, we have the slope (-4) and a point on the line (5, f(5)).

Using the point-slope form of a line, the equation of the normal line is y - f(5) = -4(x - 5).

Simplifying this equation, we get y - f(5) = -4x + 20.

Therefore, the equation of the normal line of f(x)=(x-2)/(x-3)^2 at x=5 is y = -4x + 20 + f(5).