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Worked Examples · Example 15

Q.Find the maximum and minimum values of ff, if any, of the function given by f(x)=∣x∣, x∈Rf(x) = |x|,\ x \in \mathbb{R}.

Madhya Pradesh MpbseTextbookSubjective· 2mImportance★★★★★
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Figure 6.9
Figure 6.9

The function f(x)=∣x∣f(x) = |x| is V-shaped with a sharp corner at x=0x=0. It has an absolute minimum value of 00 at x=0x=0, but no maximum value because it grows without bound as x→±∞x \to \pm\infty.

Concept and Intuition

The absolute value function is one of the simplest piecewise-defined functions. Its graph is two rays meeting at the origin: for x≥0x \ge 0, f(x)=xf(x) = x (a line with slope +1+1), and for x<0x < 0, f(x)=−xf(x) = -x (a line with slope −1-1). The key feature is the sharp corner at x=0x=0 — the function is not differentiable there, but it is continuous everywhere.

When we talk about "maximum and minimum values" of a function on its entire domain (here, all real numbers), we are looking for global (absolute) extrema. A function can have:

  • A global minimum at a point where f(x)f(x) is the smallest value over the whole domain.
  • A global maximum at a point where f(x)f(x) is the largest value over the whole domain.

For f(x)=∣x∣f(x) = |x|, the value is always non-negative. The smallest possible value is 00, achieved at x=0x=0. But can it ever be the largest? No — because as you move farther from zero in either direction, ∣x∣|x| keeps increasing without any upper bound.

Watch out

A common mistake is to think that because ∣x∣|x| has a "corner" at x=0x=0, it cannot have a minimum there. In fact, a function can have a global extremum at a point where it is not differentiable — the derivative test is sufficient but not necessary. The definition of a minimum is purely about values: f(0)≤f(x)f(0) \le f(x) for all xx, which holds here.

Tip

For functions defined on R\mathbb{R} (the whole real line), a global maximum exists only if the function is bounded above. Since ∣x∣|x| is unbounded above, no maximum exists. Similarly, a global minimum exists only if the function is bounded below and attains that lower bound — here it does, at 00.

Step-by-Step Solution

  1. Understand the domain and range.

    The domain is R\mathbb{R}, all real numbers. The range of ∣x∣|x| is [0,∞)[0, \infty) — every non-negative real number appears as an output.

  2. Check for a global minimum.

    For any x∈Rx \in \mathbb{R}, we have ∣x∣≥0|x| \ge 0. The equality ∣x∣=0|x| = 0 holds exactly when x=0x = 0.

    Therefore, f(0)=0f(0) = 0 is less than or equal to every other function value.

    So 00 is the global minimum value, attained at x=0x = 0.

  3. Check for a global maximum.

    Suppose there were a global maximum value MM. Then for all xx, ∣x∣≤M|x| \le M. But pick x=M+1x = M+1 (if M≥0M \ge 0) or x=1x = 1 (if M<0M < 0, which is impossible since ∣x∣≥0|x| \ge 0). Then ∣M+1∣=M+1>M|M+1| = M+1 > M, a contradiction.

    More simply: as x→∞x \to \infty, ∣x∣→∞|x| \to \infty, so no finite upper bound exists.

    Hence, no global maximum exists.

  4. Summarise the findings.

    • Minimum value: 00 (at x=0x=0).
    • Maximum value: none.

For f(x)=∣x∣f(x) = |x| on R\mathbb{R}:

min⁡x∈Rf(x)=0,max⁡x∈Rf(x) does not exist.\min_{x \in \mathbb{R}} f(x) = 0, \quad \max_{x \in \mathbb{R}} f(x) \text{ does not exist.}

✓Final answer

The function f(x)=∣x∣f(x) = |x| has a minimum value of 0\boxed{0} at x=0x = 0, and no maximum value.

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