Let 𝑓(𝑥) = x/x+2 . Determine the value(s) of x such that 𝑓′(𝑥) = 1.

Answer 1

First use the quotient rule to find f'(x):

dy/dx = [v'*u - u'*v] / v^2

where:

u = x

v = x+2

So that f'(x) = [1*(x+2) - 1*x] / (x+2)^2

After simplifying, f'(x) looks like:

(x - x + 2) / (x + 2)^2 --> 2 / (x^2 + 4x + 4)

Then we set that equal to 1 and solve for x. That gives us:

2 = x^2 + 4x + 4

0 = x^2 + 4x + 2

From here you can either graph it or complete the square. By doing the latter you get:

0 = (x+2)^2 - 2

2 = (x+2)^2

±√(2) = x+2 *DON'T FORGET THAT SQUARE ROOTS HAVE 2 SOLUTIONS

Final Answer: x = -2 - √(2) & -2 + √(2)

You could skip several steps here, but I just wanted to show everything.

Answer 2

f(x) = x/(x+2) = 1

x=x+2

their is no real solution

as x approaches infinity, x/(x+2) approaches 1

there are two surreal solutions

x= infinity and x=-infinity

UNLESS you meant f'(x)= 1. It's hard to read.

then

f(x) = x/(x+2)

f'(x) = [(x+2) - x]/(x+2)^2 = 1

(2)/(x^2+4x +4) = 1

2 = x^2 +4x +4

x^2 +4x +2 = 0

x = -2 +/- .5sqr(16-8)

x =-2+sqr2 or -2-sqr2