Question 27 of 33intermediate🔧 ApplyShort Answer2 marks

Find the domain and the range of the real function ff defined by f(x)=(x−1)f(x) = \sqrt{(x - 1)}.

Correct Answer

To find the domain of the real function f(x)=x−1f(x) = \sqrt{x-1}, we must determine the values of xx for which the function is defined. The function is defined only when the expression inside the square root is non-negative. Therefore, we require x−1≥0x - 1 \geq 0.

Solving the inequality x−1≥0x - 1 \geq 0, we get x≥1x \geq 1. Thus, the domain of the function is the set of all real numbers greater than or equal to 1, represented as [1,∞)[1, \infty).

For the range, since the square root of a non-negative number is always non-negative, the minimum value of f(x)f(x) is 0 (occurring at x=1x=1). As xx increases, f(x)f(x) increases without bound. Hence, the range is [0,∞)[0, \infty).

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 4 | (Page 17)
For More Understanding

Explanation

The solution applies the concept that a function is defined only for specific inputs (domain) and produces specific outputs (range), as illustrated in the provided examples (e.g., Example 21 where domain excludes undefined points, and the Modulus function section which uses the term 'non-negative'). For square root functions, the radicand must be non-negative for the output to be a real number.

Solution Steps

  1. Step 1: Set the condition for the function to be defined: the term inside the square root must be non-negative, i.e., x−1≥0x - 1 \geq 0.

  2. Step 2: Solve the inequality x−1≥0x - 1 \geq 0 to find the domain, which gives x≥1x \geq 1. Domain is [1,∞)[1, \infty).

  3. Step 3: Determine the range by observing that x−1≥0\sqrt{x-1} \geq 0 for all xx in the domain. The minimum value is 0 at x=1x=1. Range is [0,∞)[0, \infty).

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