Find the value for which f(x) = x³ -8 on [3, 7] satisfies the Mean-Value Theorem.

Question

Quizplus · Accepted Answer

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