Question 28 of 35intermediate🔧 ApplyShort Answer3 marks

Find the domain and the range of the real function ff defined by f(x)=∣x−1∣f(x) = |x - 1|.

Correct Answer

Domain: The function f(x) = |x-1| is defined for all real numbers x, because the absolute value (modulus) is defined for every real number. Hence, the domain is R (the set of all real numbers).

Range: For any real x, |x-1| ≥\geq 0. Moreover, for any non-negative real number y, we can choose x = y + 1, then f(x) = |(y+1)-1| = |y| = y. Thus, every non-negative real number is attained. Therefore, the range is the set of all non-negative real numbers, i.e., [0, ∞).

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 5 | (Page 17)
For More Understanding

Explanation

The context defines the modulus function f(x)=|x| with domain R and notes that for x≥\geq0, f(x)=x, while for x<0, f(x)=-x, implying f(x)≥\geq0. Hence its range is all non-negative reals. For f(x)=|x-1|, the same reasoning applies: the expression inside the modulus is defined for all real x, so domain = R. The output is always ≥\geq0, and by setting x = y+1 (or x = 1-y) we obtain any y≥\geq0, confirming the range is [0,∞). The context also states that a real function has subsets of R as domain and range, which is consistent.

Solution Steps

  1. Step 1: Recognize that f(x)=|x-1| involves the modulus function, which is defined for all real numbers (as per the context). Therefore, domain = R.

  2. Step 2: Note that the modulus always gives a non-negative value, so f(x) ≥\geq 0 for all x.

  3. Step 3: Show that every non-negative real y is attainable: choose x = y+1, then f(x)=|(y+1)-1|=|y|=y.

  4. Step 4: Conclude that the range is the set of all non-negative real numbers, i.e., [0, ∞).

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