f(x) = x+e^x.To show that the function is bijective.

The function f(x) is defined for every x real. So all the codomain of f(x) are images of some x. So the function is onto.

The finction is also one one as diffrerent element of domain of x has diffrent images. If not , let a and b be the different values of x forwhich f(a) = f(b). Then

a+e^a = b +e^b. Or

a-b = e^b-a^a.

(a-b) = (1+b+b^2/2!+b^3/3!+.....) -(1+1+a+a^2/2!+a^^3/3!+...)

(a-b) = (b-a)+(b^2-a^2)/2!+(b^3-a^3)/3!+.....

0 = (b-a) {(b+a)+..}

Which is possible only when b =a.

So e^x + x can not take the same values for diffrent values of the domain of x.

Therefore e^x+x is one-one.

So e^x +x is both onto and one- one. Therefore x+e^x is bijective.

To prove that a function is bijective, we have to demonstrate that the function is one-one and on-to function.

In order to verify if the function is one-one, we'll apply the first derivative test. This test verifies if the function is strictly increasing. If the function is strictly increasing, then the function is an one-one function.

So, let's calculate the first derivative:

f'(x)=1+e^x

1>0 and e^x>0, so, 1+e^x>0 => so the function f(x) is strictly increasing.

Now, we'll verify if the function is on-to function. Because f is a sum of elementary function, f is a continuous function => f is an on-to function.

We've verified that f is both, one-one and on-to function, so f(x) is a bijective function.