Which of the following statements is/are true
1. If f'(c)=0 then f has a local maximum of minimum at x=c
2. 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]
3.The Mean Value theorem can be applied to f(x)=1/x^2 on the interval [1,1]
I think 3 is right,but I'm not sure about the others.
The Options are:
1 only
2 only
1 and 3 only
1 and 2 only
Steve chillll with the typos bro shut up for once and just answer the question
Let's evaluate each statement one by one:
1. If f'(c) = 0, then f has a local maximum or minimum at x = c.
This statement is true. According to the first derivative test, if a function f'(x) changes from positive to negative at x = c, then f has a local maximum at x = c. On the other hand, if f'(x) changes from negative to positive at x = c, then f has a local minimum at x = c.
2. 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. The mean value theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), and if f'(x) = 0 for all x in (a, b), then f is constant on [a, b].
3. The Mean Value Theorem can be applied to f(x) = 1/x^2 on the interval [1, 1].
This statement is not true. The Mean Value Theorem is applicable if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b). In this case, the function f(x) = 1/x^2 is not continuous on the closed interval [1, 1] because it is not defined at x = 0. Therefore, the Mean Value Theorem cannot be applied to this function on the interval [1, 1].
Based on our evaluation, the correct statement(s) are:
- Statement 1 only
To determine which statements are true, let's analyze each one separately:
1. If f'(c) = 0, then f has a local maximum or minimum at x = c.
This statement is known as the First Derivative Test. According to the test, if f'(c) = 0 and the sign of f'(x) changes from positive to negative as x approaches c from the left, then f has a local maximum at x = c. If the sign changes from negative to positive, then f has a local minimum at x = c. Therefore, statement 1 is true.
2. If f is continuous on [a, b] and differentiable on (a, b) with f'(x) = 0 on (a, b), then f is constant on [a, b].
This statement is known as the Rolle's Theorem. According to the theorem, if f is continuous on [a, b] and differentiable on (a, b), and f(a) = f(b), then there exists at least one point c in (a, b) where f'(c) = 0. Therefore, statement 2 is true.
3. The Mean Value Theorem can be applied to f(x) = 1/x^2 on the interval [1, 1].
The Mean Value Theorem 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 instantaneous rate of change (given by f'(c)) is equal to the average rate of change (given by (f(b) - f(a))/(b - a)). In this case, the function f(x) = 1/x^2 is continuous on the interval [1, 1], but it is not differentiable on the open interval (1, 1) because it is not defined there (division by zero). Therefore, statement 3 is false.
Based on the explanations above, the correct answer is:
1 and 2 only.
1. Think of f(x) = x^3
2. If f'=0, f does not change.
3. think of f(0). It clearly does not satisfy the MVT on [-1,1]
The typos do not help...