Show that the function $f: R \rightarrow R: f(x)=x^{4}$ is neither one-one nor onto.
To prove: function is neither one-one nor onto
Given: $f: R \rightarrow R: f(x)=x^{4}$
We have,
$f(x)=x^{4}$
For, $f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow x_{1}^{4}=x_{2}^{4}$
$\Rightarrow\left(x_{1}^{4}-x_{2}^{4}\right)=0$
$\Rightarrow\left(x_{1}^{2}-x_{2}^{2}\right)\left(x_{1}^{2}+x_{2}^{2}\right)=0$
$\Rightarrow\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)\left(x_{1}^{2}+x_{2}^{2}\right)=0$
$\Rightarrow x_{1}=x_{2}$ or, $x_{1}=-x_{2}$ or, $x_{1}^{2}=-x_{2}^{2}$
We are getting more than one value of $x_{1}$ (no unique image)
$\therefore f(x)$ is not one-one
$f(x)=x^{4}$
Let $f(x)=y$ such that $y \in R$
$\Rightarrow y=x^{4}$
$\Rightarrow x=\sqrt[4]{y}$
If $y=-2$, as $y \in R$
Then $x$ will be undefined as we can't place the negative value under the square root
Hence $f(x)$ is not onto
Hence Proved
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