If F(X)=2x+1/x+2 then what is the inverse function. please explain how

y = 2x+1/x+2

To find the
inverse function:

Solution:

y
= 2x+1/x+2. Multiply by
x.

yx=2x^2+1+2x.

2x^2+(2-y)y+2.

x1
= {(y-2))+sqrt[(y-2)^2-4]}/2*2

x1 = {(2-y) -
sqrt((y-4)y)}/4

x2 = {(2-y)
-sqrt(y(y-2)y}/4.

Therefore , there is no inverse funtion
as y has the same value for two different values of x = x1 and x2. Thus   set of x and
the set of y f(x) is not a bijection.

Also y' =
(2x^2+1/x+2)' = 0 gives 2-1/x^2=0 giving x1 = 1/sqrt2 and x2 =
-1/sqrt

y"(x1) = -1/x1^2 = -1/sqrt2
<0

y"(x2) = - (-1/sqrt2)= 1sqrt. >
0

So y has minimum 2(1/sqrt1)+1/(1/sqrt2) +2 =  2(1+sqrt2)
at x = 1/sqrt2.

Y has a maximum  of
2/(-sqrt2)+1/(-1/sqrt2)+2 = 2(1-sqrt2) at x= -1/sqrt2.

At x
= 0 , y is not defined,

At x = 0, y(x+1/x) has avertical
asynptote,

As Lt x--> 0+  tha y value is
inf.

As Lt x--> 0- the y value is -
inf.

 2x+2 is the straight line which is an oblique
asymptote( as x --> infinity  or x -->
infinity).

So there are four domains where the function is
bijective.

(i) When x >=1/sqrt2,   y= 2x+1/x+2 is
bijective with inverse function:

 x = {(y-2)+sqrt[(y-2)^2
-8]}/2*2

(ii) when 0 < x < =
1/sqrt2

x = {(y-2)+sqrt[(y-2)^2-8] }
/4

(iii) when  -1/sqrtx < = x <
0.

x = {(y-2) -
sqrt[(y-2)^2-8]}/4

(iV) When x < =
-1/sqrt2.

 x= {(y-2) - sqrt{y-2)^2
-8}/4.

y has no value  in the open interval  ] 2(1-sqrt2  ,
2(1+sqrt2)  [ .