Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = 2x^2 − 5x + 1, [0, 2]
If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).
To determine if the function satisfies the hypotheses of the Mean Value Theorem (MVT) on the given interval [0, 2], we need to check two conditions:
1. Continuity: The function must be continuous on the closed interval [0, 2].
2. Differentiability: The function must be differentiable on the open interval (0, 2).
Let's check each condition one by one:
1. Continuity:
To check continuity, we need to ensure that the function is defined for all x in [0, 2] and that there are no jumps, holes, or vertical asymptotes.
The function f(x) = 2x^2 − 5x + 1 is a polynomial, and polynomials are continuous for all real numbers. Therefore, the function is continuous on the closed interval [0, 2].
2. Differentiability:
To check differentiability, we need to ensure that the function has a derivative for all x in (0, 2). In other words, the function must not have any sharp corners, cusps, or vertical tangents.
The function f(x) = 2x^2 − 5x + 1 is a polynomial, and all polynomials are differentiable for all real numbers. Therefore, the function is differentiable on the open interval (0, 2).
Since the function satisfies both conditions of the Mean Value Theorem, we can apply it. The Mean Value Theorem states that if a function is continuous on a closed interval and differentiable on the open interval, there exists at least one number c in the open interval such that f'(c) = (f(b) - f(a))/(b - a), where a and b are the endpoints of the interval.
In our case, a = 0 and b = 2. Let's find f'(x) first:
f(x) = 2x^2 − 5x + 1
Differentiating, we get:
f'(x) = 4x - 5
Now, we can find c by setting f'(c) equal to the average rate of change between f(a) and f(b):
f'(c) = (f(2) - f(0))/(2 - 0)
Substituting the values, we get:
4c - 5 = ((2)^2 - 5(2) + 1 - (0)^2 - 5(0) + 1)/2
Simplifying further:
4c - 5 = (4 - 10 + 1 - 0 + 0 + 1)/2
4c - 5 = -4/2
4c - 5 = -2
4c = 3
c = 3/4
Therefore, the only number c that satisfies the conclusion of the Mean Value Theorem is c = 3/4.
To summarize:
The function f(x) = 2x^2 − 5x + 1 satisfies the hypotheses of the Mean Value Theorem on the interval [0, 2], and the number c that satisfies the conclusion of the theorem is c = 3/4.
To determine if the function satisfies the hypotheses of the Mean Value Theorem on the given interval [0, 2], we need to check if the function is continuous on the interval [0, 2] and differentiable on the open interval (0, 2).
Step 1: Continuity
To check if the function is continuous on the interval [0, 2], we need to ensure that there are no breaks, jumps, or holes in the graph within that interval.
The given function f(x) = 2x^2 − 5x + 1 is a polynomial function. Polynomial functions are continuous for all real numbers.
Therefore, the function is continuous on the interval [0, 2].
Step 2: Differentiability
To check if the function is differentiable on the open interval (0, 2), we need to ensure that the derivative exists and is defined for all x-values in the interval (0, 2).
The derivative of the given function f(x) = 2x^2 − 5x + 1 can be found by applying the power rule:
f'(x) = 4x - 5
The derivative is defined for all real numbers, so it is defined for all x-values in the interval (0, 2).
Therefore, the function is differentiable on the open interval (0, 2).
Conclusion:
The given function f(x) = 2x^2 − 5x + 1 satisfies the hypotheses of the Mean Value Theorem on the interval [0, 2].
To find all numbers c that satisfy the conclusion of the Mean Value Theorem, we need to find the x-value(s) at which the derivative is equal to the average rate of change of the function on the interval [0, 2].
The average rate of change, or the slope of the secant line, is given by:
m = (f(2) - f(0))/(2 - 0)
m = (2(2)^2 - 5(2) + 1 - (2(0)^2 - 5(0) + 1))/(2 - 0)
m = (8 - 10 + 1 - 1)/(2 - 0)
m = -2/2
m = -1
To satisfy the conclusion of the Mean Value Theorem, there must be at least one number c in the interval (0, 2) such that f'(c) = -1.
To find the x-value(s) at which the derivative is equal to -1, we can set the derivative equation equal to -1 and solve for x:
4x - 5 = -1
4x = 4
x = 1
Therefore, the number c that satisfies the conclusion of the Mean Value Theorem is c = 1.
Final answer: The function satisfies the hypotheses of the Mean Value Theorem on the interval [0, 2], and the number that satisfies the conclusion of the Mean Value Theorem is c = 1.
f(x) is differentiable on the interval. So, you know there is a c such that
f'(c) = (f(2)-f(0)/(2-0) = (-1-2)/2 = -3/2
So, just solve for c in
4c-5 = -3/2
c = 7/8
as a check, view
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x^2-5x%2B1+and+y+%3D+%28-3%2F2%29%28x-7%2F8%29+-+59%2F32