Hi all,

can anyone help to solve this?

Let f: R -> R defined as f(x) = 9X2. Determine whether or not the function is one-to-one and onto.

Thanks in Advance

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- Nov 1st 2007, 06:04 AM #1

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- Nov 1st 2007, 06:53 AM #2
it is not 1-1 nor onto..

not 1-1:

by def:

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

take f(x) = 9, then x = 1 or x = -1.

therefore not 1-1.

not onto:

$\displaystyle f: R \rightarrow R$

for every element in the range, there must be an element in domain..

take -1 in range, does there an element in the domain that will give you -1 when mapped using your function?

- Nov 2nd 2007, 03:13 AM #3

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- Nov 2nd 2007, 04:09 AM #4
the definitions themselves:

**Def:**

A function $\displaystyle f:R \rightarrow R$ is said to be**one-to-one**if $\displaystyle f(x)=f(y) $, then $\displaystyle x=y $.

okay, let's take your function $\displaystyle f(x)=9x^2$ as an example.

so, suppose $\displaystyle f(x)=f(y)$

then $\displaystyle 9x^2 = 9y^2 \implies x^2=y^2 \implies x=|y|$

here, x is either y or -y which contradicts the definition. hence, $\displaystyle f(x)=9x^2$ is not 1-1.

another explanation is that, any value in the range corresponds to a unique value in the domain..

**Def:**

A function $\displaystyle f:X \rightarrow Y$ is said to be**onto Y**if the range of f is Y.

in your function, $\displaystyle f:R \rightarrow R$ defined by $\displaystyle f(x)=9x^2$, the range is $\displaystyle [0, +\infty)$ and not**R**, therefore, f is not**onto R**.

here, it says that for**all**the values in the range must corresponds to a value in the domain.