Which of the following statements is/are true?
I.If f '(c) = 0, then f has a local maximum or minimum at x = c.
II.If f is continuous on [a, b] and differentiable on (a, b) and f '(x) = 0 on (a, b), then f is constant on [a, b].
III.The Mean Value Theorem can be applied to f(x) = 1/x^2 on the interval [-1, 1].
I only
II only
I and III only
I and II only
iii cannot be true at x=0 The function is not continous, which is a requirement for the MVT.
So would that mean that it is only the first one?
The correct answer is II only.
Explanation:
I. If f'(c) = 0, it means that the derivative of f at point c is equal to zero. This indicates that the function may have a local maximum, local minimum, or a flat point at x = c. However, it is not guaranteed that the function has a local maximum or minimum at x = c. Therefore, statement I is not always true.
II. If f is continuous on [a, b] and differentiable on (a, b), and f'(x) = 0 on (a, b), then f is constant on [a, b]. This statement is known as the constant value theorem. When the derivative of a function is zero on a given interval, it implies that the function has a constant value on that interval. Therefore, statement II is true.
III. The Mean Value Theorem (MVT) states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) where the derivative of f at c is equal to the average rate of change of f over the interval [a, b]. However, for the function f(x) = 1/x^2, the function is not continuous at x = 0 since it is not defined for x = 0. Hence, the MVT cannot be applied to f(x) = 1/x^2 on the interval [-1, 1]. Therefore, statement III is not true.
Therefore, the correct answer is II only.
To determine which of the statements are true, let's analyze each statement individually.
I. If f '(c) = 0, then f has a local maximum or minimum at x = c.
This statement is true and is a direct consequence of the First Derivative Test. If a function f is differentiable at a point c, and f '(c) = 0, then it means the graph of f has a horizontal tangent at x = c. At that point, f can have a local maximum or minimum depending on the behavior of the function on either side of c.
II. If f is continuous on [a, b] and differentiable on (a, b) and f '(x) = 0 on (a, b), then f is constant on [a, b].
This statement is also true and is known as the Constant Function Theorem. If a function f is continuous on the closed interval [a, b] and its derivative f '(x) is identically zero on the open interval (a, b), then f is constant on the entire closed interval [a, b].
III. The Mean Value Theorem can be applied to f(x) = 1/x^2 on the interval [-1, 1].
To determine if the Mean Value Theorem can be applied to a function, we need to check two conditions: continuity on the closed interval and differentiability on the open interval. In this case, the function f(x) = 1/x^2 is not continuous at x = 0 because it is not defined at that point, and therefore, the Mean Value Theorem cannot be applied to it. Hence, statement III is false.
Therefore, the correct answer is I and II only.