Question

Check the injectivity and subjectivity of;

1 ) If f(x)=x²+x+1.find fog,gof,gog,fof​

Answer 1

(i)

f(x)=x

2

It is seen that for x,y∈N,

f(x)=f(y)⇒x

2

=y

2

⇒x=y

∴f is injective

Now, 2∈N, But, there does not exist any x

in N such that f(x)=x

2

=2.

∴f is not surjective.

Hence, function f is injective but not surjective.

(ii)

f(x)=x

2

It is seen that f(−1)=f(1)=1, but −1



=1

∴f is not injective.

Now, −2∈Z. But, there does not exist any

element x∈Z such that f(x)=x

2

=−2

∴f is not surjective.

Hence, function f is neither injective nor surjective.

(iii)

f(x)=x

2

It is seen that f(−1)=f(1)=1, but −1



=1.

∴f is not injective.

Now, −2∈R. But, there does not exist any element x∈R such that f(x)=x

2

=−2

∴f is not surjective.

Hence, function f is neither injective nor surjective.

(iv)

f(x)=x

3

It is seen that for x,y∈N,f(x)=f(y)⇒x

3

=y

3

⇒x=y

∴f is injective.

Now, 2∈N. But, there does not exist any element

x in domain N such that f(x)=x

3

=2

∴f is not surjective.

Hence, function f is injective but not surjective.

(v)

f(x)=x

3

It is seen that for x,y∈Z,f(x)=f(y)⇒x

3

=y

3

⇒x=y

∴f is injective.

Now, 2∈Z. But, there does not exist any element

x in domain Z such that f(x)=x

3

=2

∴f is not surjective.

Hence, function f is injective but not surjective.

I hope this is ur question and if my answer is wrong sry for tht.