# Let $f:R \to R{\text{ and g : R}} \to {\text{R}}$be defined by $f(x) = {x^2}{\text{ and g(x) = x + 1}}$.Show that $f(g(x)) \ne g(f(x))$

Last updated date: 25th Mar 2023

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Answer

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Hint: Find the composition of one function in terms of another function.

Now $f(x)$and $g(x)$is given to us such that $f(x) = {x^2}{\text{ and g(x) = x + 1}}$

Now $f:R \to R{\text{ and g : R}} \to {\text{R}}$is given as it means that the domain of f is R and so as the range, similarly the domain of g is R and its range is also R.

We have to show that $f(g(x)) \ne g(f(x))$

So let’s first compute $f(g(x))$

$ \Rightarrow f(x + 1)$

$ \Rightarrow {(x + 1)^2}$……………………………………….. (1)

Now let’s compute $g(f(x))$

$ \Rightarrow g({x^2})$

$ \Rightarrow {x^2} + 1$…………………………………………. (2)

Clearly equation 1 is not equal to equation 2 as ${(x + 1)^2} \ne {x^2} + 1$

Hence $f(g(x)) \ne g(f(x))$ proved.

Note: Whenever we come across such problems the only key concept that will be involved is how to find the composition of one function into another and it’s been explained above. Some questions may involve some tricky parts as they may not be having the same domain and range, so pay special attention to this part before solving.

Now $f(x)$and $g(x)$is given to us such that $f(x) = {x^2}{\text{ and g(x) = x + 1}}$

Now $f:R \to R{\text{ and g : R}} \to {\text{R}}$is given as it means that the domain of f is R and so as the range, similarly the domain of g is R and its range is also R.

We have to show that $f(g(x)) \ne g(f(x))$

So let’s first compute $f(g(x))$

$ \Rightarrow f(x + 1)$

$ \Rightarrow {(x + 1)^2}$……………………………………….. (1)

Now let’s compute $g(f(x))$

$ \Rightarrow g({x^2})$

$ \Rightarrow {x^2} + 1$…………………………………………. (2)

Clearly equation 1 is not equal to equation 2 as ${(x + 1)^2} \ne {x^2} + 1$

Hence $f(g(x)) \ne g(f(x))$ proved.

Note: Whenever we come across such problems the only key concept that will be involved is how to find the composition of one function into another and it’s been explained above. Some questions may involve some tricky parts as they may not be having the same domain and range, so pay special attention to this part before solving.

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