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

Let y = (2x + 1)/(x+2)

Multiply by (x+2)

==> y(x+2) = 2x + 1

Open brackets:

==> yx + 2y = 2x + 1

Group terms with x on the left sides:

==> yx -2x = 1-2y

Now factor x:

==> x (y-2) = 1-2y

Now divide by (y-2)

==> x= (1-2y)/(y-2)

Then the inverse is :

**F^-1 (x) = (1-2x)/(x-2)**

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) [ .

A function means 3 things: domain of definition, range of values of function and the analytic expression of the function.

A function is invertible if and only if is bijective, that means that it has to be both, injective and surjective.

To check if it's injective, we'll differentiate the function and we'll verify if the resulted expression is strictly positive.

F'(x) = [(2x+1) / (x+2)]'

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

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

F'(x) = (2x + 4 - 2x - 1)/(x+2)^2

We'll eliminate like terms:

F'(x) = 3/(x+2)^2

Since the denominator is always positive, for any value of x, F'(x) is strictly positive, so F'(x)>0 => F(x) injective.

To prove that F(x) is a surjective function, we'll have to prove that F(x) is a continuous function and it is.

So, F(x) is a bijective function, so it's invertible.

Now, we'll calculate the inverse:

Let's put F(x) = y

So, y = (2x+1)/(x+2)

We'll cross multiply and we'll get:

(x+2)*y = 2x + 1

We'll remove the brackets:

x*y + 2y = 2x + 1

We'll move all terms in x to the left side and all terms without the variable x, to te right side(including the terms in y but without x).

x*y - 2x = 1 - 2y

We'll factorize by x, to the left side:

x*(y - 2) = 1 - 2y

We'll divide by (y-2):

x = (1-2y)/(y-2)

Let's note the inverse function of F(x) as f(x).

**The inverse function is: f(y) = (1-2y)/(y-2).**