Question 27 of 35intermediate🔧 ApplyShort Answer3 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

Domain: For the function f(x)=x−1f(x) = \sqrt{x-1} to be defined as a real function, the expression under the square root must be non-negative.

x−1≥0x - 1 \geq 0 x≥1x \geq 1

Therefore, Domain = {x:x∈R,x≥1}\{x : x \in \mathbf{R}, x \geq 1\} or [1,∞)[1, \infty).

Range: Since x≥1x \geq 1, we have x−1≥0x - 1 \geq 0, which means x−1≥0\sqrt{x-1} \geq 0.

The minimum value occurs at x=1x = 1, giving f(1)=1−1=0f(1) = \sqrt{1-1} = 0. As xx increases beyond 1, f(x)f(x) increases without bound.

Therefore, Range = {y:y≥0}\{y : y \geq 0\} or [0,∞)[0, \infty).

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

Explanation

This question tests understanding of domain and range for square root functions. The domain is found by ensuring the radicand (expression under square root) is non-negative, since square roots of negative numbers are not real. Setting x - 1 ≥\geq 0 gives x ≥\geq 1. For range, since the output of a square root function is always non-negative and can grow arbitrarily large, the range is all non-negative real numbers [0, ∞).

Solution Steps

  1. Step 1: For domain, set the expression under square root ≥\geq 0: x - 1 ≥\geq 0

  2. Step 2: Solve: x ≥\geq 1, so Domain = [1, ∞)

  3. Step 3: For range, note that x−1\sqrt{x-1} ≥\geq 0 for all x ≥\geq 1

  4. Step 4: Minimum value is 0 (at x = 1), maximum is unbounded, so Range = [0, ∞)

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