B
Rolle's Theorem
I and III are false - consider (x^2-1)^2 + 2/3 on [√(1 - 1/√3),√(1 + 1/√3)]
f" = 0 at x = 1/√3, but that is outside [a,b]
I. f(x) = 0
II. f'(x) = 0
III. f"(x) = 0
a. I only
b. II only
c. I and II only
d. I, II, and III
Rolle's Theorem
I and III are false - consider (x^2-1)^2 + 2/3 on [√(1 - 1/√3),√(1 + 1/√3)]
f" = 0 at x = 1/√3, but that is outside [a,b]
So, I shifted it up until 1/√3 was outside [a,b]
If f(a) = f(b) = 1, it means that at both points a and b, the function evaluates to 1. But does that mean it must equal 0 or have a derivative of 0 or even a second derivative of 0 at some point between a and b?
Let's consider each option:
I. f(x) = 0: Aha! This one is easy. If f(a) = f(b) = 1 and we're asking if f(x) = 0 at some point, then the answer would be no. The function doesn't need to equal 0 at any point between a and b, so that option is out.
II. f'(x) = 0: Ah, the derivative! Well, if f(a) = f(b) = 1, it doesn't mean that the derivative of f(x) must equal 0 at some point. So, option II is also out.
III. f"(x) = 0: The second derivative! Now we're getting fancy. But hold your horses, because this option is also incorrect. Just because f(a) = f(b) = 1, it doesn't imply that the second derivative of f(x) must equal 0 at some point between a and b.
So, after considering all the options, the answer is none of the above. None of the given options must be true for at least one value of x between a and b when f(a) = f(b) = 1.
Hope that cleared things up, even if it wasn't the punchline you were hoping for!
Rolle's theorem states that for a function f(x) that satisfies the following conditions:
1. f(x) is continuous on the closed interval [a, b]
2. f(x) is differentiable on the open interval (a, b)
3. f(a) = f(b), i.e., the function takes on the same value at the endpoints
Then, there exists at least one value c between a and b such that f'(c) = 0.
In this case, we know that f(a) = f(b) = 1. Thus, the conditions of Rolle's theorem are satisfied. Therefore, option II, f'(x) = 0, must be true for at least one value of x between a and b.
Options I and III are not necessarily true. The function f(x) could either have roots or not and its second derivative f"(x) could be non-zero throughout the interval (a, b).
Therefore, the correct answer is b. II only.
The Intermediate Value Theorem states that if a continuous function f(x) takes on two different values, such as f(a) and f(b), at two points a and b in an interval, then the function must also take on every value between those two values at least once in that interval.
In this case, we know that f(a) = f(b) = 1. Therefore, by the Intermediate Value Theorem, the function f(x) must take on every value between 1 and 1 at least once between a and b.
Now, let's evaluate the given statements:
I. f(x) = 0: There is no information given about f(x) being equal to zero. So, statement I cannot be concluded.
II. f'(x) = 0: The derivative of a polynomial function of degree greater than 2 can have multiple roots or points where it equals zero. Therefore, statement II could potentially be true for at least one value of x between a and b.
III. f"(x) = 0: The second derivative of a polynomial function of degree greater than 2 can also have multiple roots or points where it equals zero. Therefore, statement III could potentially be true for at least one value of x between a and b.
In summary, statement I cannot be concluded, but both statement II and statement III could potentially be true for at least one value of x between a and b. Therefore, the correct answer is:
b. II only