Show that the function

Solution:

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