Problem 21 - A Textbook of Differential Calculus

Question

The range of the function \(f(x)=\sqrt{x-1}+2 \sqrt{3-x}\) is (a) \([\sqrt{2}, 2 \sqrt{2}\) ] (b) \([\sqrt{2}, \sqrt{10}]\) (c) \([2 \sqrt{2}, \sqrt{10}]\) (d) \([1,3]\)

Step-by-Step Solution

Answer

The range of the function is (a) \([ \sqrt{2}, 2 \sqrt{2}]\).

Step 1: Understand the Domain of the Function

The function is defined as long as the expressions under the square roots are non-negative. This gives the inequalities: \(x - 1 \geq 0\) and \(3 - x \geq 0\). Solving these gives \(x \geq 1\) and \(x \leq 3\). Hence, the domain of \(f(x)\) is \(1 \leq x \leq 3\).

Step 2: Analyze the Function

Within the domain, \(x\) ranges between 1 and 3. As \(x\) increases from 1 to 3, \(\sqrt{x-1}\) increases from 0 to \(\sqrt{2}\), while \(2 \sqrt{3-x}\) decreases from \(2 \sqrt{2}\) to 0.

Step 3: Find Extremes of the Function

At \(x = 1\), the value of \(f(x) = \sqrt{1-1} + 2 \sqrt{3-1} = 0 + 2 \times \sqrt{2} = 2 \sqrt{2}\). At \(x = 3\), \(f(x) = \sqrt{3-1} + 2 \sqrt{3-3} = \sqrt{2} + 0 = \sqrt{2}\).

Step 4: Conclude the Range of the Function

The range of \(f(x)\) as \(x\) takes all values from 1 to 3 includes all values from \(f(3) = \sqrt{2}\) to \(f(1) = 2 \sqrt{2}\). Hence, the range is \([\sqrt{2}, 2 \sqrt{2}]\), corresponding to option (a).

Key Concepts

Domain of Function

When discussing the domain of a function, we're essentially referring to the set of all possible input values (usually represented with the variable \(x\)) for which the function is mathematically defined and produces a real number result. For square root functions, the domain is constrained by the requirement that the expression inside the square root must be non-negative, as square roots of negative numbers are not real by conventional math.

• The square root function \(f(x) = \sqrt{x-1}\) is defined only when \(x-1 \geq 0\), which simplifies to \(x \geq 1\).
• Similarly, for the function \(2 \sqrt{3-x}\), we need \(3-x \geq 0\), leading to \(x \leq 3\).

Combining both conditions, the overall domain for this function is \(1 \leq x \leq 3\). This limited range of \(x\) values determines where the function can operate and produce real values.

Square Root Function

Square root functions belong to a special category of functions where the square root of an expression defines the function's output. These functions are key in understanding radicals and algebraic manipulation. For the function \(f(x) = \sqrt{x-1} + 2\sqrt{3-x}\), each term involves a square root, imposing certain behaviors:

• \(\sqrt{x-1}\) increases monotonically as \(x\) increases, starting from 0 when \(x = 1\) to reaching \(\sqrt{2}\) when \(x = 3\).
• \(2 \sqrt{3-x}\) decreases as \(x\) increases, from the maximum value of \(2\sqrt{2}\) at \(x = 1\) to 0 at \(x = 3\).

This illustrates how each square root term's behavior, determined by its argument, contributes both an increasing and decreasing trend across its respective domains, affecting the overall function's outcome in its domain.

Function Analysis

Analyzing a function involves scrutinizing its behavior across its domain, identifying key values such as minima, maxima, and intercepts. For the function \(f(x) = \sqrt{x-1} + 2\sqrt{3-x}\), initiated from the domain \([1, 3]\), an analysis reveals how the sum of the two contrasting behaviors define the range. The square root function \(\sqrt{x-1}\) increases from 0 to \(\sqrt{2}\) whereas \(2\sqrt{3-x}\) decreases from \(2\sqrt{2}\) to 0. Thus:

• At \(x = 1\), \(f(x)\) is at \(2\sqrt{2}\).
• At \(x = 3\), \(f(x)\) approaches \(\sqrt{2}\).

Drawing together their contributions, we conclude the range of \(f(x)\) from the lowest value \(\sqrt{2}\) at \(x = 3\) to \(2\sqrt{2}\) at \(x = 1\). Delineating this span ensures complete understanding of the function's behavior, confirming the range as \([\sqrt{2}, 2\sqrt{2}]\). This analysis demonstrates the synergy of increasing and decreasing trends in forming the function's overall trajectory.