Find the value for which f(x) = x² + 5 on [1, 5] satisfies the Mean-Value Theorem. A) 1.5 B) 3 C) 4.5 D) 2.33 E) 3.66

Question

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The Mean Value Theorem (MVT) states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that f'(c) = [f(b) - f(a)] / (b - a). Given f(x) = x^2 + 5, we find f'(x) = 2x. We need to find c such that f'(c) = [f(5) - f(1)] / (5 - 1). Calculating, f(5) = 5^2 + 5 = 30, f(1) = 1^2 + 5 = 6, so [f(5) - f(1)] / (5 - 1) = (30 - 6) / 4 = 24 / 4 = 6. Setting f'(c) = 2c = 6 gives c = 3, which is in the interval [1, 5].

Find the Value for Which F(x) = X2 + 5

Find the Value for Which F(x) = X² + 5