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Which of these functions from R to R are bijections?

  • $f(x)=-3x+4$
  • $f(x)=-3x^2+7$
  • $f(x)=\dfrac{x+1}{x+2}$
  • $f(x)=x^5+1$

This question aims to identify the bijective functions from the given list of functions.

In mathematics, functions are the foundation of calculus representing various kinds of relationships.  A function is a rule, expression, or law that specifies an association between a variable known as an independent variable and a dependent variable. This implies that if $f$ is a function and with a set of potential inputs usually known as the domain, will map an element, say $x$, from the domain to specifically one element, say $f(x)$, in the set of potential outputs called the co-domain of the function.

A bijective function is also called a bijection, invertible function, or one-to-one correspondence. This is a type of function which is responsible for assigning specifically one element of a set to exactly one element of another set and vice-versa. In this type of function, every element of both sets is paired with each other in such a way that no element of both sets remain unpaired. Mathematically, let $f$ be a function, $y$ be any element in its co-domain, then there must one and only one element $x$ such that $f(x)=y$.

Expert Answer

$f(x)=-3x+4$ is bijective. To prove that, let:

$f(y)=-3y+4$

$f(x)=f(y)$

$-3x+4=-3y+4$ or $x=y$

which means that $f(x)$ is one-one.

Also, let $y=-3x+4$

$x=\dfrac{4-y}{3}$

or $f^{-1}(x)=\dfrac{4-x}{3}$

So, $f(x)$ is onto. Since $f(x)$ is both one-to-one and surjective, therefore, it is a bijective function.

$f(x)=-3x^2+7$ is not a bijective function being quadratic, since $f(-x)=f(x)$.

$f(x)=\dfrac{x+1}{x+2}$ fails to be a bijective function since it is undefined at $x=-2$. But the condition for a function to be bijective from $R\to R$ is that it should be defined for every element of $R$.

$f(x)=x^5+1$ is bijective. To prove that let:

$f(y)=y^5+1$

$f(x)=f(y)$

$x^5+1=y^5+1$ or $x=y$

which means that $f(x)$ is one-one.

Also, let $y=x^5+1$

$x=(y-1)^{1/5}$

or $f^{-1}(x)=(x-1)^{1/5}$

So $f(x)$ is onto. Since $f(x)$ is both one-to-one and surjective, therefore, it is a bijective function.

Example

Prove that $f(x)=x+1$ is a bijective function from $R\to R$.

Solution

To prove that the given function is bijective, first prove that it is both one-to-one and an onto function.

Let $f(y)=y+1$

For a function to be one-to-one:

$f(x)=f(y)$ $\implies x=y$

$x+1=y+1$

$x=y$

For a function to be onto:

Let  $y=x+1$

$x=y-1$

$f^{-1}(x)=x-1$

Since $f(x)$ is one-to-one and onto, this implies that it is bijective.

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