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Question:

Let f, g, and h be real-valued functions defined on the interval [0, 1] by f(x) = ex² + e-x², g(x) = xex² + e-x², and h(x) = x²ex² + e-x². If a, b, and c denote, respectively, the absolute maximum of f, g, and h on [0, 1], then what is the relationship between a, b, and c?

a=b=c

a≠b and c≠b

a=b and c≠b

a=c and a≠b

Solution:

  1. ex² ≥ x ex² ≥ x²ex² ∀x ∈ [0,1]
  2. ex² ≥ xex² ∀x ∈ [0,1]
    ex² ≥ x ex²
    1 ≥ x
    This is true for all x ∈ [0,1].
  3. x ex² ≥ x²ex² ∀x ∈ [0,1]
    x ex² ≥ x²ex²
    x ≥ x²
    This is true for all x ∈ [0,1].
    Therefore, ex² ≥ xex² ≥ x²ex² ∀x ∈ [0,1]
    Adding e-x² to all sides:
    ex² + e-x² ≥ xex² + e-x² ≥ x²ex² + e-x²
    i.e. f(x) ≥ g(x) ≥ h(x) ∀x ∈ [0,1]
    Equality holds when x = 1
    i.e. f(x) ≥ g(x) ≥ h(x) ∀x ∈ [0,1]
    Thus, f(1) is the greatest.
    Thus, a = b = c = e + 1/e
    Hence, a = b = c.