How do you find the points where the graph of the function #y=(4x)/(x+1)# has horizontal tangents and what is the equation?

Question

Evelyn Agard · Accepted Answer

Find the derivative, set it to #0#, solve the equation, then find the #y# value(s) at the solution(s). The horizontal tangent at that point has equation #y=# the value.

#y = (4x)/(x+1)# has derivative #y' = 4/(x+1)^2# which is never #0#, so there is no horizontal tangent line. Finding #y'# Use the quotient rule: #y' = ((4)(x+1) - (4x)(1))/(x+1)^2 = (4x+4-4x)/(x+1)^2 = 4/(x+1)^2# A different example #y = x^3-3x^2+5# #y' = 3x^2-6x = 0# at #x=0# and at #x=2# At #x=0#, we get #y=5# so the line tangent to the curve at #(0,5)# has equation #y=5#. At #x=2#, we get #y=1# so the line tangent to the curve at #(2,1)# has equation #y = 1#.

Christopher Agresta · Answer

undefined To find the points where the graph of the function y=(4x)/(x+1) has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of the function y=(4x)/(x+1) using the quotient rule. The derivative is given by:

dy/dx = [(4)(x+1) - (4x)(1)] / (x+1)^2

Next, we set the derivative equal to zero and solve for x:

[(4)(x+1) - (4x)(1)] / (x+1)^2 = 0

Simplifying the equation, we have:

4(x+1) - 4x = 0

Expanding and combining like terms:

4x + 4 - 4x = 0

4 = 0

Since the equation 4 = 0 is not true, there are no values of x that satisfy this equation. Therefore, the graph of the function y=(4x)/(x+1) does not have any points where it has horizontal tangents.