Bounds Math Example 2

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Example 2

hard
Show that for all real \(x\), \(x^2 \geq 0\). What is the greatest lower bound (infimum) of \(x^2\)?

Solution

  1. 1
    Case 1: \(x > 0\). Then \(x^2 = x \times x > 0\).
  2. 2
    Case 2: \(x = 0\). Then \(x^2 = 0\).
  3. 3
    Case 3: \(x < 0\). Then \(x^2 = (-|x|)^2 = |x|^2 > 0\).
  4. 4
    In all cases \(x^2 \geq 0\).
  5. 5
    The greatest lower bound is \(0\), achieved at \(x=0\).

Answer

\(x^2 \geq 0\) for all \(x\); infimum = 0
Squaring any real number gives a non-negative result. The minimum (infimum) of \(x^2\) is 0, achieved when \(x=0\).

About Bounds

The upper and lower limits within which a quantity must lie; often expressed as a≤x≤ba \leq x \leq b.

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More Bounds Examples

Example 1 medium

For (f(x) = -x^2 + 6x - 5), find the maximum value (upper bound) on ([0, 5]).

Example 3 medium

For (g(x) = 3x + 1) on ([0, 4]), find the minimum and maximum values.

Example 4 hard

Prove that (sin(x) leq 1) and (sin(x) geq -1) for all (x) using the unit circle definition.

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