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Find the value of c guaranteed by the Mean Value Theorem (MVT) for f ( x ) =√ (81 − x²) over the interval [ 0 , 9 ].

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Final answer:

The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the interval (a, b) where the instantaneous rate of change (slope) is equal to the average rate of change over the interval. In this problem, the value of c guaranteed by the MVT for f(x) = √(81 - x²) over the interval [0, 9] is approximately 6.36.

Step-by-step explanation:

The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the interval (a, b) where the instantaneous rate of change (slope) is equal to the average rate of change over the interval.

In the given problem, we have f(x) = √(81 - x²) and the interval [0, 9].

To find the value of c guaranteed by the MVT, we first need to find the derivative of f(x):

f'(x) = -x/√(81 - x²)

Next, we find the average rate of change over the interval [0, 9]:

The average rate of change = (f(9) - f(0))/(9 - 0) = (√(81 - 81) - √(81 - 0))/(9 - 0) = (0 - 9)/9 = -1

Now, we set the derivative equal to the average rate of change and solve for x:

-x/√(81 - x²) = -1

Simplifying the equation, we get:

x/√(81 - x²) = 1

Squaring both sides of the equation:

x²/(81 - x²) = 1

Multiplying both sides by (81 - x²):

x² = 81 - x²

Bringing like terms to one side:

2x² = 81

Dividing both sides by 2:

x² = 40.5

Taking the square root of both sides:

x = ±√40.5

Since x must be in the interval [0, 9], we only consider the positive square root:

x = √40.5 ≈ 6.36

Therefore, the value of c guaranteed by the MVT is approximately 6.36.

User Rob Kwasowski
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