Given f: R → R is a function defined by
![\[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{4}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{7}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-50aa06b787b4af5959bf67d4c806ab1a_l3.png "Rendered by QuickLaTeX.com")
Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
⇒
![\[4{{x}^{3~}}+\text{ }7~=~4{{y}^{3~}}+~7\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-38d03c0a03b295ec14a762a8db1b0043_l3.png "Rendered by QuickLaTeX.com")
⇒
![\[4{{x}^{3~}}=~4{{y}^{3}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-334c5f5e5a582089a03801db37d61557_l3.png "Rendered by QuickLaTeX.com")
⇒
![\[{{x}^{3~}}=~{{y}^{3}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-43f59610d1603053b8981aea377e91d4_l3.png "Rendered by QuickLaTeX.com")
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain)
f(x) = y
⇒
![\[4{{x}^{3~}}+\text{ }7~=~y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b7c34d4b5907e8f145ea6288ce61d990_l3.png "Rendered by QuickLaTeX.com")
⇒
![\[4{{x}^{3~}}=~y~-\text{ }7\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5b0718c5182523eeca97c00b0ed409ef_l3.png "Rendered by QuickLaTeX.com")
⇒
![\[{{x}^{3}}~=\text{ }\left( y\text{ }\text{ }7 \right)/4\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-afa531eb847c9ca20fae0d7f0aaf5fa5_l3.png "Rendered by QuickLaTeX.com")
⇒
![\[x\text{ }=\sqrt[3]{\left( y-7 \right)/4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-923e684727db52b374a873b9cec6247c_l3.png "Rendered by QuickLaTeX.com")
in R
So, for every element in the co-domain, there exists some pre-image in the domain. f is onto.
Since, f is both one-to-one and onto, it is a bijection.