Question

If f(x) =x² and g(x) = x+1.show that gof not equal to fog​

Answer 1

Step-by-step explanation:

Given,

$f(x) = {x}^{2}$

$g(x) = x + 1$

To find gof and fog

We know that,

gof = g(f(x))

Therefore, we will get,

$= > g(f(x)) = {x}^{2} + 1$

And, we know that,

fog = f(g(x))

Therefore, we will get,

$= > f(g(x)) = {(x + 1)}^{2} \\ \\ = > f(g(x)) = {x}^{2} + 2x + 1$

Clearly, we have,

${x}^{2} + 1 \ne {x}^{2} + 2x + 1$

Therefore, we will get,

$g(f(x)) \ne f(g(x))$

Thus, we get,

gof ≠ fog

Hence, proved.