Question

Let f : R → R and g : R → R where R is the set of Real Numbers. Find FOG and GOF where f(x) = x^2 - 2 and g(x) = x + 4. State where these functions are Injective, Surjective, Bijective. ​

Answer 1

SOLUTION

GIVEN

Let f : R → R and g : R → R where R is the set of Real Numbers

where f(x) = x² - 2 and g(x) = x + 4

TO DETERMINE

• f o g , g o f

• State where these functions are Injective, Surjective, Bijective.

EVALUATION

Here it is given that f : R → R and g : R → R where R is the set of Real Numbers

where f(x) = x² - 2 and g(x) = x + 4

Now

$\sf{(f \circ g)(x)}$

$\sf{ = f (g(x))}$

$\sf{ = f (x + 4)}$

$\sf{ = {(x + 4)}^{2} - 2}$

$\sf{ = {x}^{2} + 8x + 14}$

$\boxed { \: \: \sf{ (f \circ g)(x)= {x}^{2} + 8x + 14} \: \: }$

Again

$\sf{(g \circ f)(x)}$

$\sf{ = g ( f(x))}$

$\sf{ = g ( {x}^{2} - 2)}$

$\sf{ = {x}^{2} - 2 + 4}$

$\sf{ = {x}^{2} + 2}$

$\boxed{ \: \: \sf{ (g \circ f)(x)= {x}^{2} + 2} \: \: }$

CHECKING FOR f o g

INJECTIVE

$\sf{ (f \circ g)(x)= {x}^{2} + 8x + 14}$

Here the function is real values

Discriminant = 8 > 0

So both roots are real

Suppose the roots are a and b then

$\sf{ (f \circ g)(a)= (f \circ g)(b) = 0 \: \: but \: \: a \ne \: b}$

So the function is not injective

SURJECTIVE

Clearly - 20 in the codomain set R has no pre image in the domain set R

So the function is not surjective

Thus f o g is not bijective

CHECKING FOR g o f

INJECTIVE

Here the given function is

$\sf{ (g \circ f)(x)= {x}^{2} + 2}$

Clearly

$\sf{ (g \circ f)( - 2)= (g \circ f)( 2) = 6 \: \: but \: - 2 \ne \: 2}$

So the function is not injective

SURJECTIVE

Here we observe that

$\sf{ (g \circ f)( x) \geqslant 2 \: \: \: for \: all \: x}$

Clearly - 20 in the codomain set R has no pre image in the domain set R

So the function is not surjective

Thus g o f is not bijective

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