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We actually need f to be differentiable on (0, 4) as well. However, we can take the condition 2 ≤ f '(x) ≤ 5 to imply this.

This is a simple application of the Mean Value Theorem. Since f is continuous on [0, 4] and differentiable on (0, 4), the MVT says that there is a value z ∈ (0, 4) such that

f '(z) = [f(4) - f(0)] / (4 - 0). In other words, f(4) = f(0) + 4 f '(z) = 1 + 4 f '(z).

Since 2 ≤ f '(z) ≤ 5, 9 ≤ 1 + 4 f '(z) ≤ 21. So 9 ≤ f(4) ≤ 21.