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 number c in (a, b) such that:
f'(c) = (f(b) – f(a)) / (b – a)
To apply the MVT to the function f(x) = x on the interval [0, 9], we will first check the conditions of the theorem:
- The function f(x) = x is continuous everywhere, including the interval [0,9].
- The function f(x) = x is differentiable everywhere, including on (0,9).
Now we can find f(a) and f(b):
- Let a = 0 and b = 9. Therefore, f(0) = 0 and f(9) = 9.
Using the MVT formula, we calculate:
f'(c) = (f(9) – f(0)) / (9 – 0) = (9 – 0) / (9 – 0) = 9 / 9 = 1.
Next, we find f'(x) to get the derivative:
f'(x) = 1.
Now we solve for c:
Since f'(c) = 1, we find that c can be any value in the interval (0, 9), as the derivative of the function is constant at 1 across this entire range. Therefore, the conclusion of the Mean Value Theorem is satisfied for any c in that interval.
In conclusion, any number c within the interval (0, 9) satisfies the Mean Value Theorem for the given function.