Show that the function f(x)=4x^3−15x^2+9x+8
satisfies the three hypotheses of Rolle’s theorem on the interval [0,3]. Then find the values of c on the interval [0,3] that are guaranteed by Rolle’s theorem. Give your answer as a set of values, e.g., {1,3,5,6,7}
To show that the function f(x) = 4x^3 - 15x^2 + 9x + 8 satisfies the three hypotheses of Rolle’s theorem on the interval [0, 3], we need to verify the following conditions:
1. f(x) must be continuous on the closed interval [0, 3]:
To check the continuity of f(x), we need to ensure that there are no holes, jumps, or asymptotes within the interval [0, 3]. The function f(x) is a polynomial, and polynomials are continuous for all values of x. Hence, f(x) = 4x^3 - 15x^2 + 9x + 8 is continuous on [0, 3].
2. f(x) must be differentiable on the open interval (0, 3):
To verify the differentiability of f(x), we need to make sure that the derivative of f(x) exists for all values of x in the open interval (0, 3). The derivative of f(x) is f'(x) = 12x^2 - 30x + 9. Since f'(x) is a polynomial, it exists and is defined for all values of x. Therefore, f(x) is differentiable on (0, 3).
3. f(0) = f(3):
To check if f(0) is equal to f(3), we evaluate the function at these points:
f(0) = 4(0)^3 - 15(0)^2 + 9(0) + 8 = 8
f(3) = 4(3)^3 - 15(3)^2 + 9(3) + 8 = 8
Since f(0) = f(3) = 8, the third condition of Rolle's theorem is satisfied.
Now, we can find the values of c on the interval [0, 3] that are guaranteed by Rolle's theorem. These values of c correspond to the points where the derivative f'(x) of f(x) is zero (since the second condition of Rolle's theorem is satisfied).
Setting f'(x) = 0, we solve the equation 12x^2 - 30x + 9 = 0. This quadratic equation can be factored as (2x - 3)(6x - 3) = 0.
Setting each factor equal to zero, we have:
2x - 3 = 0 => x = 3/2
6x - 3 = 0 => x = 1/2
Therefore, the values of c on the interval [0, 3] that are guaranteed by Rolle's theorem are c = {1/2, 3/2}.
To verify if a function satisfies the three hypotheses of Rolle's theorem on the interval [0, 3], we need to check the following:
1. Continuity: The function must be continuous on the closed interval [0, 3].
2. Differentiability: The function must be differentiable on the open interval (0, 3).
3. Equal function values: The function must have equal function values at the endpoints.
Let's go through each hypothesis one by one:
1. Continuity: To determine if the function is continuous on the interval [0, 3], we need to check if it is continuous at every point within that interval. Since the given function is a polynomial, it is continuous for all real numbers. Therefore, it is continuous on [0, 3].
2. Differentiability: To determine if the function is differentiable on the open interval (0, 3), we need to check if its derivative exists and is continuous within that interval. Taking the derivative of f(x), we get:
f'(x) = 12x^2 - 30x + 9
The derivative is also a polynomial, which means it is continuous for all real numbers. Therefore, f(x) is differentiable on (0, 3).
3. Equal function values: To satisfy this hypothesis, we need to show that f(0) = f(3). Evaluating the function at x = 0 and x = 3, we have:
f(0) = 4(0)^3 - 15(0)^2 + 9(0) + 8 = 8
f(3) = 4(3)^3 - 15(3)^2 + 9(3) + 8 = 91
Since f(0) = 8 is equal to f(3) = 91, the function has equal function values at the endpoints.
Having verified all three hypotheses, we can assert that Rolle's theorem applies to the function f(x) on the interval [0, 3].
To determine the values of c guaranteed by Rolle's theorem, we find the critical points within the interval (0, 3). Critical points correspond to values of x where the derivative is zero or undefined.
Setting f'(x) = 0, we have:
12x^2 - 30x + 9 = 0
Solving this quadratic equation, we find two solutions: x = 1/3 and x = 3/2.
Since these critical points are within the interval (0, 3), we can conclude that Rolle's theorem guarantees the existence of values c in the interval (0, 3) where f'(c) = 0.
Therefore, the values of c guaranteed by Rolle's theorem are {1/3, 3/2}.