[Solution] Function Pre.

Domain and Range of a Function¶

✅ Simple Definition (Not Bookish)¶

A function takes an input, does something to it, and gives you an output.

The domain is all the possible values you are allowed to give as input.
The range is all the values you get back as output.

You can think of a function like a vending machine.

The buttons you can press (like A1, B2, C3) — that’s the domain.
The snacks that come out — that’s the range.

🎯 Formal But Simple Definition:¶

Domain of a function is the complete set of all possible input values (x) for which the function is defined.
Range of a function is the complete set of all possible output values (f(x)) that the function can give.

🧠 Example:¶

Let’s say we have a function:
f(x) = x²

Step 1: Domain¶

You can put any real number into x (positive, negative, or zero), and you will get a valid result.

✅ So, the domain is: All real numbers
In symbols: x ∈ ℝ

Step 2: Range¶

No matter what number you put, the output will always be zero or positive (because square of any real number is ≥ 0).

✅ So, the range is: All real numbers greater than or equal to 0
In symbols: f(x) ≥ 0 or f(x) ∈ [0, ∞)

Example 2:¶

Given the function:

\[ f(x) = 3x^2 + x + 2 \]

Find the domain and range of the function.

✅ Domain of f(x):¶

The function \(f(x) = 3x^2 + x + 2\) is a polynomial function.
Polynomial functions are defined for all real numbers.

So,

\[ \text{Domain } D_f = \{x \in \mathbb{R}\} = \mathbb{R} \]

✅ Range of f(x):¶

Let,

\[ y = f(x) = 3x^2 + x + 2 \tag{1} \]

Now, rearrange (1) to form a quadratic in \(x\):

\[ 3x^2 + x + (2 - y) = 0 \]

This is a quadratic equation in \(x\).
For real values of \(x\), the discriminant must be ≥ 0.

Discriminant formula:

\[ D = b^2 - 4ac \]

In our case:

\(a = 3\),
\(b = 1\),
\(c = 2 - y\)

So,

\[ D = 1^2 - 4 \cdot 3 \cdot (2 - y) = 1 - 12(2 - y) = 1 - 24 + 12y = 12y - 23 \]

To get real \(x\), we need:

\[ 12y - 23 \geq 0 \Rightarrow y \geq \frac{23}{12} \]

✅ Range:¶

So, the range of the function is:

\[ R_f = \left[ \frac{23}{12}, \infty \right) \]

📌 Final Answer:¶

Domain: \(\mathbb{R}\)
Range: \(\left[ \frac{23}{12}, \infty \right)\)

\[ f(x) = \sqrt{x - 1} + \sqrt{5 - x}, \quad \text{for } x \in [1, 5] \]

✅ Step 1: Understand the Function's Behavior¶

This function is the sum of two square roots, and their domains are opposite in nature:

\(\sqrt{x - 1}\) increases as \(x\) increases.
\(\sqrt{5 - x}\) decreases as \(x\) increases.

So when \(x = 1\), the first term is minimum and the second term is maximum.
When \(x = 5\), the first term is maximum and the second term is minimum.

✅ Step 2: Find Minimum and Maximum Values¶

👉 At \(x = 1\):¶

\[ f(1) = \sqrt{1 - 1} + \sqrt{5 - 1} = \sqrt{0} + \sqrt{4} = 0 + 2 = 2 \]

👉 At \(x = 5\):¶

\[ f(5) = \sqrt{5 - 1} + \sqrt{5 - 5} = \sqrt{4} + \sqrt{0} = 2 + 0 = 2 \]

👉 At \(x = 3\) (Midpoint for symmetry):¶

\[ f(3) = \sqrt{3 - 1} + \sqrt{5 - 3} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \approx 2.828 \]

This is the maximum value.

✅ Step 3: Final Answer¶

Minimum value: 2
Maximum value: \(2\sqrt{2} \approx 2.828\)

📌 Final Range:¶

\[ \boxed{\text{Range } = [2, 2\sqrt{2}]} \]