This problem can be solved using the Mean Value Theorem (MVT). The Mean Value Theorem states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in the interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].

First, we can apply the MVT to the interval [0, 1]. The function f is continuous on [0, 1] and differentiable on (0, 1), and f(1) - f(0) = 3 - 0 = 3. So, the average rate of change of f on [0, 1] is 3 / (1 - 0) = 3. By the MVT, there exists some c1 in (0, 1) such that f'(c1) = 3.

Next, we can apply the MVT to the interval [1, 4]. The function f is continuous on [1, 4] and differentiable on (1, 4), and f(4) - f(1) = -3 - 3 = -6. So, the average rate of change of f on [1, 4] is -6 / (4 - 1) = -2. By the MVT, there exists some c2 in (1, 4) such that f'(c2) = -2.

Now, consider the function f' on the interval [c1, c2]. By the Intermediate Value Theorem (IVT), since f'(c1) = 3 and f'(c2) = -2 and f' is continuous (because f is differentiable), there must exist some c in (c1, c2) ⊆ (0, 4) such that f'(c) = -1. This is the c that the problem is asking for.

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This problem can be solved using the Mean Value Theorem (MVT). The Mean Value Theorem states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in the interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].

First, we can apply the MVT to the interval [0, 1]. The function f is continuous on [0, 1] and differentiable on (0, 1), and f(1) - f(0) = 3 - 0 = 3. So, the average rate of change of f on [0, 1] is 3 / (1 - 0) = 3. By the MVT, there exists some c1 in (0, 1) such that f'(c1) = 3.

Next, we can apply the MVT to the interval [1, 4]. The function f is continuous on [1, 4] and differentiable on (1, 4), and f(4) - f(1) = -3 - 3 = -6. So, the average rate of change of f on [1, 4] is -6 / (4 - 1) = -2. By the MVT, there exists some c2 in (1, 4) such that f'(c2) = -2.

Now, consider the function f' on the interval [c1, c2]. By the Intermediate Value Theorem (IVT), since f'(c1) = 3 and f'(c2) = -2 and f' is continuous (because f is differentiable), there must exist some c in (c1, c2) ⊆ (0, 4) such that f'(c) = -1. This is the c that the problem is asking for.

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This problem can be solved using the Mean Value Theorem (MVT). The Mean Value Theorem states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in the interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].

First, we can apply the MVT to the interval [0, 1]. The function f is continuous on [0, 1] and differentiable on (0, 1), and f(1) - f(0) = 3 - 0 = 3. So, the average rate of change of f on [0, 1] is 3 / (1 - 0) = 3. By the MVT, there exists some c1 in (0, 1) such that f'(c1) = 3.

Next, we can apply the MVT to the interval [1, 4]. The function f is continuous on [1, 4] and differentiable on (1, 4), and f(4) - f(1) = -3 - 3 = -6. So, the average rate of change of f on [1, 4] is -6 / (4 - 1) = -2. By the MVT, there exists some c2 in (1, 4) such that f'(c2) = -2.

Now, consider the function f' on the interval [c1, c2]. By the Intermediate Value Theorem (IVT), since f'(c1) = 3 and f'(c2) = -2 and f' is continuous (because f is differentiable), there must exist some c in (c1, c2) ⊆ (0, 4) such that f'(c) = -1. This is the c that the problem is asking for.

Suppose that f : [0, 4] → R is continuous on [0, 4], is differentiable on (0, 4) and thatf (0) = 0, f (1) = 3, f (4) = −3.Show that there exists c ∈ (0, 4) such that f ′(c) = −1

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