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

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Jim H
Apr 7, 2016

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.

Explanation:

#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#.

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