What is the equation of the tangent line of #f(x) =sqrt((x^2-1)/(x-5))# at # x = 1#?

Question

Luke Alvarez · Accepted Answer

#x = 1#.

Finding the point of contact's y-coordinate is the first thing we do. #f(1) = sqrt((1^2 - 1)/(1 - 5))# #f(1) = sqrt(0/5)# #f(1) = 0# Now, we apply the chain and quotient rules to differentiate the function. We call #u = (x^2 - 1)/(x - 5)#. #u' = (2x(x- 5) - 1(x^2- 1))/(x- 5)^2# #u' = (2x^2 - 10x - x^2 + 1)/(x - 5)^2# #u' = (x^2 - 10x + 1)/(x- 5)^2# We now let #y = sqrt(u)# and #u# being as noted above. The derivative is given by #dy/dx= dy/(du) xx (du)/dx#. #dy/dx = 1/(2u^(1/2)) xx (x^2 - 10x + 1)/(x - 5)^2# #dy/dx = (x^2 - 10x + 1)/(2((x^2 - 1)/(x - 5))^(1/2)(x- 5)^2)# The derivative represents the instantaneous rate of change, or slope , at any given point #x = a#. We want to find the slope at #x = 1#. #m_"tangent" = (1^2- 10(1) + 1)/(2((1^2 -1)/(1 - 5))^(1/2)(1 - 5)^2)# #m_"tangent" = (-8)/(0)# #m_"tangent" = O/# Since the slope is undefined, the tangent is vertical. Therefore, the equation is of the form #x =a#, where #(a, b)# is the point of contact. Hence, the equation is #x= 1#. I hope this is useful!

Paisley Johnson · Answer

undefined The equation of the tangent line of f(x) = sqrt((x^2-1)/(x-5)) at x = 1 is y = -2x + 3.