Let g(x)=x^(3)-16 x and let c be the number that satisfies the Mean Value Theorem for g on the interval [-4,2]. What is c ? Choose 1 answer: (A) -3 (B) -2 (C) 0 (D) 1

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Let  g(x)=x^(3)-16 x and let  c be the number that satisfies the Mean Value Theorem for  g on the interval  [-4,2]. What is  c ? Choose 1 answer: (A) -3 (B) -2 (C) 0 (D) 1

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Check Conditions: To apply the Mean Value Theorem (MVT), we need to ensure that the function g(x) is continuous on the closed interval [-4, 2] and differentiable on the open interval (-4, 2). The function g(x) = x^3 - 16x is a polynomial, which is continuous and differentiable everywhere. Therefore, it satisfies the conditions for the MVT.Apply MVT: The MVT states that there exists at least one number c in the open interval (-4, 2) such that g'(c) is equal to the average rate of change of the function over the interval [-4, 2]. The average rate of change is given by [g(b) - g(a)] / (b - a), where a = -4 and b = 2.Calculate g(-4) and g(2): First, we calculate g(-4) and g(2). g(-4) = (-4)^3 - 16(-4) = -64 + 64 = 0, and g(2) = (2)^3 - 16(2) = 8 - 32 = -24.Calculate Average Rate of Change: Now, we calculate the average rate of change: [g(2) - g(-4)] / (2 - (-4)) = (-24 - 0) / (2 + 4) = -24 / 6 = -4.Find g'(x): Next, we find g'(x), which is the derivative of g(x) with respect to x. g'(x) = d/dx [x^3 - 16x] = 3x^2 - 16.Set up MVT equation: According to the MVT, there exists a number c such that g'(c) = -4. So we set the derivative equal to -4 and solve for c: 3c^2 - 16 = -4.Solve for c: We rearrange the equation to solve for c: 3c^2 - 16 + 4 = 0, which simplifies to 3c^2 - 12 = 0.Factor Quadratic Equation: Divide both sides by 3 to simplify further: c^2 - 4 = 0.Check Solutions: Now we factor the quadratic equation: (c - 2)(c + 2) = 0.Final Result: Setting each factor equal to zero gives us two possible solutions for c: c - 2 = 0 or c + 2 = 0, which means c = 2 or c = -2.However, we must check which of these solutions lies within the open interval (-4, 2). The number c = 2 does not lie within the interval (-4, 2), but c = -2 does. Therefore, c = -2 is the number that satisfies the Mean Value Theorem for g(x) on the interval [-4, 2].

Let $g(x)=x^{3}-16 x$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $[-4,2]$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-3$ $\newline$ (B) $-2$ $\newline$ (C) $0$ $\newline$ (D) $1$

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Let g(x)=x3−16x g(x)=x^{3}-16 x g(x)=x3−16x and let c c c be the number that satisfies the Mean Value Theorem for g g g on the interval [−4,2] [-4,2] [−4,2].\newlineWhat is c c c ?\newlineChoose 111 answer:\newline(A) −3-3−3\newline(B) −2-2−2\newline(C) 000\newline(D) 111

Let $g(x)=x^{3}-16 x$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $[-4,2]$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-3$ $\newline$ (B) $-2$ $\newline$ (C) $0$ $\newline$ (D) $1$