Question

For the bjective functions determine their inverses.
Il Find fog and goj, where f(x) = 1 + 1 and g() =+2 are functions from R to R​

Answer 1

Answer:

ogj

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given, function f:R→R such that f(x)=1+x

2

,

Let A and B be two sets of real numbers.

Let x

1

,x

2

∈A such that f(x

1

)=f(x

2

).

⇒1+x

1

2

=1+x

2

2

⇒x

1

2

−x

2

2

=0⇒(x

1

−x

2

)(x

1

+x

2

)=0

⇒x

1

=±x

2

. Thus f(x

1

)=f(x

2

) does not imply that x

1

=x

2

.

For instance, f(1)=f(−1)=2, i.e. , two elements (1, -1) of A have the same image in B. So, f is many-one function.

Now, y=1+x

2

⇒x=

y−1

⇒elements < y have no pre-image in A (for instance an element -2 in the codomain has no pre-image in the domain A). So, f is not onto.

Hence, f is neither one-one onto. So, it is not bijective.