maximum value of the expression?

Question

The maximum value of the expression ( for all real values of x) is:

x^2 -2x+4
---------------
x^2+2x+4

1) 4
2)2
3)3
4)No fixed maximum value

Answer 1

d/dx = [(x² + 2x + 4)*(2x-2) - (x² - 2x + 4)*(2x+2)]/(x²+2x+4)²
set d/dx = 0

(x² + 2x + 4)*(2x-2) - (x² - 2x + 4)*(2x+2) = 0
2x³ + 4x² + 8x - 2x² - 4x - 8 -(2x³ - 4x² + 8x + 2x² - 4x + 8) = 0
2x³ + 2x² + 4x - 8 - 2x³ + 2x² - 4x - 8 = 0
4x² - 16 = 0
x² = 4
x = ± 2

Check value at 2 and -2
@2 value is 1/3
@-2 value is 3

Max value is 3.

This is a total max not just a local max because as x→∞ f(x) = 1 and x→-∞ f(x) = -1. The previous answer of no max is wrong.

Answer 2

derivating
y´=1/(x^2+2x+4)^2 *[(x^2+2x+4)(2x-2)-(x^2-2x+4)(2x+2)]=
=4/(x^2+2x+4)^2*[x^2-4] The sign of y´is

++++++ -2 -------- +2++++++++ so
x=-2 is a relative maximum
f(-2)=3 is the absolute maximum as lim y x=>+-infinity is 1

Answer 3

Let y =

x^2 -2x+4
---------------
x^2+2x+4

Do a differentiation dy/dx and set dy/dx = 0 for max or min.
you will find that there is no maximum.