Let f(x)=αx^2+βx+γ be a quadratic function, so α≠0, and let I=[a,b].
a) Check f satisfies the hypothesis of the Mean Value Theorem.
b)Show that the number c ∈ (a,b) in the Mean Value Theorem is the midpoint of the interval I.
every polynomial is continuous and differentiable everywhere.
At the midpoint, x = (a+b)/2
f'(x) = 2αx+β
So, show that f'((a+b)/2) = (f(b)-f(a)/(b-a)
a) In order for the function f to satisfy the hypothesis of the Mean Value Theorem, it must meet two conditions: continuity on the closed interval [a, b] and differentiability on the open interval (a, b).
1. Continuity: For f to be continuous on the closed interval [a, b], it must be continuous at every point within the interval, including the endpoints a and b.
2. Differentiability: For f to be differentiable on the open interval (a, b), the derivative f'(x) must exist for every value of x within the interval (a, b).
To check the first condition, we can examine continuity at the endpoints:
At x = a:
f(a) = αa^2 + βa + γ
At x = b:
f(b) = αb^2 + βb + γ
Both f(a) and f(b) are polynomial functions, which are continuous at all points. Therefore, f is continuous on [a, b].
Now let's check the differentiability condition. We need to find the derivative f'(x) and verify its existence on (a, b):
f'(x) = 2αx + β
The derivative f'(x) of a quadratic function exists for all values of x. Hence, f is differentiable on (a, b).
In conclusion, the function f(x) satisfies the hypothesis of the Mean Value Theorem.
b) According to the Mean Value Theorem, there exists a number c in the interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
Using the derivative we found earlier, f'(c) = 2αc + β.
Applying the Mean Value Theorem equation, we have:
2αc + β = (f(b) - f(a))/(b - a)
Expanding f(b) and f(a) using the given quadratic function, we get:
2αc + β = (αb^2 + βb + γ - αa^2 - βa - γ)/(b - a)
Simplifying further, we have:
2αc + β = (αb^2 - αa^2 + βb - βa)/(b - a)
Factoring out α and β from the numerator, we get:
2αc + β = α(b^2 - a^2)/(b - a) + β(b - a)/(b - a)
Simplifying again, we have:
2αc + β = α(b + a) + β
As α ≠ 0, we can divide the expression by 2α, which gives:
c = (b + a)/2
Therefore, c is the midpoint of the interval I = [a, b].
In conclusion, we have shown that the number c ∈ (a, b) in the Mean Value Theorem is the midpoint of the interval I.
a) To check if the function f satisfies the hypothesis of the Mean Value Theorem, we need to verify two conditions:
1. Continuity: The function f(x) must be continuous on the closed interval [a, b]. This means that f(x) should not have any breaks, jumps, or undefined points within the interval.
2. Differentiability: The function f(x) must be differentiable on the open interval (a, b). This means that the derivative of f(x) exists for all x in the interval (a, b).
Continuity: To check continuity, we need to ensure that there are no jumps or breaks in the function f(x) within the interval [a, b]. Since f(x) is a quadratic function, it is a polynomial, and all polynomials are continuous on their entire domain. Therefore, f(x) is continuous on [a, b] by default.
Differentiability: To check differentiability, we need to verify that the derivative of f(x) exists for all x in the interval (a, b). The derivative of f(x) with respect to x is denoted as f'(x) and is equal to 2αx + β.
Since α≠0, the derivative 2αx + β is a linear function, and linear functions are differentiable on their entire domain. Thus, f(x) is differentiable on (a, b).
Therefore, f satisfies the hypothesis of the Mean Value Theorem.
b) The Mean Value Theorem states that if f is continuous on [a, b] and differentiable on (a, b), then there exists at least one number c in the interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In this case, f'(x) = 2αx + β. To find the number c ∈ (a, b) in the Mean Value Theorem, we need to solve the equation:
2αc + β = (f(b) - f(a))/(b - a)
Now let's find f(b) and f(a):
f(b) = αb^2 + βb + γ
f(a) = αa^2 + βa + γ
Substituting these values back into the equation, we have:
2αc + β = (αb^2 + βb + γ - αa^2 - βa - γ)/(b - a)
Canceling out γ, we simplify the equation:
2αc + β = (αb^2 + βb - αa^2 - βa)/(b - a)
Next, we can factor out common terms from the numerator:
2αc + β = α(b^2 - a^2)/(b - a) + β(b - a)/(b - a)
Simplifying further:
2αc + β = α(b + a) + β
Finally, solving for c:
2αc = α(b + a) + β - β
2αc = α(b + a)
c = (b + a)/2
This shows that the number c, which satisfies the Mean Value Theorem, is the midpoint of the interval I = [a, b].
Therefore, c = (b + a)/2.