Question

minimum value of (x^2+2x+4)/(x+2) ?

Answer 1

Minimum value is -∞ for x in (-∞, -2).
Minimum value is  2  for x in  (-2, ∞).

It depends on what we are looking at.

y = (x^2 +  2x + 4) / (x+2)
   =  x  +  4/(x+2)  =   -2 + (x+2) + 4/(x+2)
****  We can say that  z + 4/z  is  minimum  when  z = √4 = 2.
****   So  x+2 = 2, ie.,  at x = 0,  y is minimum.  So  Minimum value = 2.


==> Alternately:  Take derivatives on both sides:
   dy/dx  =  1  - 4 / (x+2)² = 0   for     x+2 = +2  or  -2
                      ie.,  x = 0  or  -4
   d²y/dx²  =  8 /(x+2)³ > 0 for x > -2.   Choose x such that y" > 0.

So for x = 0, y is minimum.   
Minimum value = (0^2 + 2 *0+4)/(0+2) =  2.

See the diagram.

Answer 2

Step-by-step explanation:

$let \: y = \frac{ {x}^{2} + 2x + 4 }{x + 2}$

$then...$

${x}^{2} + x(2 - y) + 2(2 - y) = 0$

$as \: x \: is \: real \: . \: discriminant \: should \: be \\ \: greater \: than \: or \: equal \: to \: 0$

${(2 - y)}^{2} - 8(2 - y) \geqslant 0$

$(2 - y)(2 - y - 8) \geqslant 0$

$(y - 2)(y + 6) \geqslant 0$

$y \leqslant - 6$

$or$

$y \geqslant 2$

Hence minimum positive value of y is 2