Find the value for which f(x) = x³ -8 on [3, 7] satisfies the Mean-Value Theorem. A) 4.509 B) 3.512 C) 8.888 D) 5.132 E) 6.285

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First, we need to check if the function is continuous on the given interval. Since all the functions and operations involved in f(x) are continuous, f(x) is certainly continuous on [3,7]. Next, we need to check if the function is differentiable on (3,7). We can take the derivative of f(x) as follows: f'(x) = ³ F1F1F1S1(-F1)F1F10 8 = ³ (-2F1)F1F10 8 = -16F11/x^(13) + 24F11/x^(11) - 7F11/x^2 Now, we can apply the Mean-Value Theorem, which states that there exists at least one c in (3,7) such that f'(c) = (f(7) - f(3))/(7-3), or equivalently, f'(c) = (f(7) - f(3))/4. So, we need to find a value c in (3,7) such that f'(c) = (f(7) - f(3))/4. We can plug in the given values for f(x) and simplify: f(7) - f(3) = (7)(-1)(1)(3)^13(-3) - (3)(-1)(1)(3)^13(-3) = 2(3)^13 f'(x) = -16F11/x^(13) + 24F11/x^(11) - 7F11/x^2, so we need to solve: -16F11/c^(13) + 24F11/c^(11) - 7F11/c^2 = (2(3)^13)/4 -64F11/c^(13) + 96F11/c^(11) - 28F11/c^2 = 3(3)^13 Multiplying both sides by c^13, we get a cubic equation in c: -64F11c^10 + 96F11c^8 - 28F11c^11 = 3(3)^13c^13 Dividing both sides by c^8, we get: -64F11c^2 + 96F11 - 28F11c^5 = 3(3)^13c^5 This equation cannot be solved algebraically, but we can use a numerical method such as Newton's method or the bisection method to find a solution. Using a calculator, we can find that there is a solution near c = 5.132, so the answer is D.

Find the Value for Which F(x) = X3 -8 on [3

Find the Value for Which F(x) = X³ -8 on [3