Let h(x)=x^(3)-9x^(2)+7x and let c be the number that satisfies the Mean Value Theorem for h on the interval -3

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Let  h(x)=x^(3)-9x^(2)+7x and let  c be the number that satisfies the Mean Value Theorem for  h on the interval  -3 <= x <= 6. What is  c ? Choose 1 answer: (A) -1 (B) 0 (C) 3 (D) 4

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Mean Value Theorem: The Mean Value Theorem 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 number c in the interval (a, b) such that f'(c) is equal to the average rate of change of the function over [a, b]. The average rate of change is given by [f(b) - f(a)] / (b - a).Calculate Endpoints: First, we need to find the values of h(x) at the endpoints of the interval. These are h(-3) and h(6). h(-3) = (-3)^3 - 9(-3)^2 + 7(-3) = -27 - 81 - 21 = -129. h(6) = (6)^3 - 9(6)^2 + 7(6) = 216 - 324 + 42 = -66.Calculate Average Rate: Now, we calculate the average rate of change of h(x) over the interval [-3, 6]. Average rate of change = [h(6) - h(-3)] / (6 - (-3)) = (-66 - (-129)) / (6 + 3) = 63 / 9 = 7.Find Derivative: Next, we find the derivative of h(x), which is h'(x). h'(x) = d/dx [x^3 - 9x^2 + 7x] = 3x^2 - 18x + 7.Apply Mean Value Theorem: According to the Mean Value Theorem, there exists a number c such that h'(c) = 7 (the average rate of change we calculated). So we set the derivative equal to 7 and solve for c: 3c^2 - 18c + 7 = 7.Set Derivative Equal: Subtract 7 from both sides to set the equation to zero: 3c^2 - 18c = 0.Factor and Solve: Factor out the common term c: c(3c - 18) = 0.Final Value of c: Set each factor equal to zero and solve for c: c = 0 or 3c - 18 = 0.If 3c - 18 = 0, then 3c = 18 and c = 6. However, c = 6 is an endpoint of the interval and cannot be the value we are looking for according to the Mean Value Theorem. Therefore, the only possible value for c within the interval (-3, 6) is c = 0.

Let $h(x)=x^{3}-9 x^{2}+7 x$ and let $c$ be the number that satisfies the Mean Value Theorem for $h$ on the interval $-3 \leq x \leq 6$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-1$ $\newline$ (B) $0$ $\newline$ (C) $3$ $\newline$ (D) $4$

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