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.