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டவுன்லோடு செய்யுங்கள்Let us look at a few key terminologies relating to functions in this section.

For that matter, let us consider the function \(f : X \rightarrow Y\).

**1. Domains and co-domains**:

In the function \(f : X \rightarrow Y\), the set \(X\) is the domain, and the set \(Y\) is the co-domain.

Domain \(=\) Set \(X\) \(=\) \(\{x_1\), \(x_2\), \(x_3\), \(x_4\), \(x_5\),\(...\}\)

Co-domain \(=\) Set \(Y\) \(=\) \(\{y_1\), \(y_2\), \(y_3\), \(y_4\), \(y_5\),\(...\}\)

**2. Images and preimages**:

If \(f(x) = y\), the image of \(x\) is '\(y\)' and the pre-image of \(y\) is '\(x\)'.

From the figure given above, we can draw the following inferences.

For the image \(y_1\), \(x_1\) is its preimage.

For the image \(y_2\), \(x_2\) is its preimage.

For the image \(y_3\), \(x_3\) is its preimage.

For the image \(y_4\), \(x_4\) is its preimage.

For the image \(y_5\), \(x_5\) is its preimage.

**3. Describing domain of a function**:

Let \(f(x)\) be \(\frac{1}{x^2 - 5x + 20}\).

The function mentioned above holds for all real numbers except for \(4\) and \(5\).

In such cases, we can write \(f(x)\) as \(\frac{1}{x^2 - 5x + 20}\), where \(x \in R - \{4, 5\}\).

**4. Conditions to be a function**:

\(f : X \rightarrow Y\) is only a function if and only if the following conditions are met:

- Every preimage of \(f\) has an image.
- Each of the images is unique.

In

**Figure 1**, each of the preimages has unique images. Hence,**Figure 1**depicts a function.Similarly,

**Figure 2**is also a function.But in

**Figure 3**, the preimage \(x_3\) has the images \(y_2\) and \(y_3\). Since a preimage can only have one unique image,**Figure 3**does not represent a function. Also, \(x_2\) does not have an image.**5. Range**:The set of images of \(f\) is the range of that function.

From the image given above, Range \(=\) \(\{y_1\), \(y_2\), \(y_3\), \(y_4\), \(y_5\}\)

Important!

Let \(n(K) = t\), and \(n(L) = s\).

Then the total number of functions between \(K\) and \(L\) is \(s^t\).

For \(f : X \rightarrow Y\), \(n(X)\) \(=\) \(2\) and \(n(Y)\) \(=\) \(3\).

The total number of elements in \(f\) \(=\) \(n(Y)^{n(X)}\) \(=\) \(3^2\)