Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = x3 + x − 9, [0, 2]
Well, the short answer is yes, because the function is continuous and continuously differentiable.
However you probably want to find a point in the interval where the tangent slope is the same as the slope of the line between end points
so
(0 , -9) to (2, 0)
slope = (0 - -9) / 2 = 4.5
where does dy/dx = 4.5 ?
dy/dx = 3 x^2 + 1
4.5 = 3 x^2 + 1
x^2 = 3.5/ 3
yes, x is between 0 and 2
Well, the Mean Value Theorem states that if a function is continuous on a 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 the function is equal to the average rate of change of the function over the interval [a, b].
In this case, the function f(x) = x^3 + x - 9 is indeed continuous on the closed interval [0, 2] and differentiable on the open interval (0, 2). Therefore, it does satisfy the hypotheses of the Mean Value Theorem on the given interval.
But you know, it's interesting to think about this function. It's like a little roller coaster ride for your equations, with the x^3 term providing the ups and downs, and the simpler x and constant terms keeping things grounded. It's a perfect blend of excitement and stability!
To determine if a function satisfies the hypotheses of the Mean Value Theorem on a given interval, we need to check if two conditions are met:
1. The function must be continuous on the closed interval [a, b].
2. The function must be differentiable on the open interval (a, b).
Let's check these conditions for the given function f(x) = x^3 + x - 9 on the interval [0, 2]:
1. Continuity: To verify continuity on the closed interval [0, 2], we need to make sure that the function is defined and has no discontinuities in this interval. The given function is a polynomial, and polynomials are continuous everywhere. Therefore, f(x) = x^3 + x - 9 is continuous on the closed interval [0, 2].
2. Differentiability: To check differentiability on the open interval (0, 2), we need to ensure that the function has a derivative at every point in this interval. The derivative of f(x) = x^3 + x - 9 is f'(x) = 3x^2 + 1. As a polynomial, f'(x) exists for all values of x, including the open interval (0, 2).
Since both conditions are satisfied, the given function f(x) = x^3 + x - 9 satisfies the hypotheses of the Mean Value Theorem on the interval [0, 2].
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 meets two conditions: continuity and differentiability.
1. Continuity:
A function is continuous on a closed interval [a, b] if it is continuous on the interval (a, b) and exists at the endpoints a and b. To check continuity, we need to evaluate the function at the endpoints [0, 2] and see if it exists.
f(0) = (0)^3 + (0) - 9 = -9
f(2) = (2)^3 + (2) - 9 = 5
Since the function exists at both endpoints, it is continuous on the closed interval [0, 2].
2. Differentiability:
A function is said to be differentiable on a closed interval [a, b] if it is differentiable on the interval (a, b) and its derivative also exists at the endpoints a and b. To check differentiability, we need to evaluate the derivative of the function and see if it exists at the endpoints [0, 2].
f'(x) = 3x^2 + 1
f'(0) = 3(0)^2 + 1 = 1
f'(2) = 3(2)^2 + 1 = 13
Since the derivative exists at both endpoints, the function is differentiable on the closed interval [0, 2].
Therefore, the function f(x) = x^3 + x - 9 satisfies the hypotheses of the Mean Value Theorem on the given interval [0, 2].