Let f(x) be a function that is differentiable everywhere and has a derivative f′(x)=4x2−4x+2 . Verify that the Intermediate Value Theorem for Derivatives applies to the function f′(x) on the interval [0,2] , and find the value of c guaranteed by the theorem such that f′(c)=5 .
To verify that the Intermediate Value Theorem for Derivatives applies to the function f′(x) on the interval [0,2], we need to show that f′(x) is continuous on [0,2]. Since f′(x) is the derivative of f(x) and f(x) is differentiable everywhere, f′(x) is continuous on its domain.
To find the value of c guaranteed by the theorem such that f′(c) = 5, we can use the Mean Value Theorem. The Mean Value Theorem states that if a function f(x) is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in (a, b) such that f′(c) = (f(b) - f(a))/(b - a).
Applying the Mean Value Theorem to f′(x) on [0, 2], we have:
f′(c) = (f(2) - f(0))/(2 - 0)
We can find f(2) using the antiderivative of f′(x):
f(x) = ∫(4x^2 - 4x + 2) dx = (4/3)x^3 - 2x^2 + 2x + C
Evaluating f(2):
f(2) = (4/3)(2)^3 - 2(2)^2 + 2(2) + C = 32/3 - 8 + 4 + C = 20/3 + C
Similarly, we can find f(0):
f(0) = (4/3)(0)^3 - 2(0)^2 + 2(0) + C = 0 + 0 + 0 + C = C
Substituting f(2) and f(0) back into the Mean Value Theorem equation, we have:
f′(c) = (20/3 + C - C)/(2 - 0) = (20/3)/(2) = 10/3
So, there exists at least one c in (0, 2) such that f′(c) = 10/3. However, we want to find a value of c such that f′(c) = 5.
Since f′(x) = 4x^2 - 4x + 2, we need 4c^2 - 4c + 2 = 5. Simplifying this equation, we get:
4c^2 - 4c - 3 = 0
Using the quadratic formula, we find that the solutions to this equation are c = (4 ± sqrt(16 + 48))/8. Simplifying further, we have:
c = (4 ± sqrt(64))/8 = (4 ± 8)/8
Therefore, the possible values for c are c = 1 or c = 3/2.
Hence, by the Intermediate Value Theorem for Derivatives, there exists at least one value of c in the interval (0, 2) such that f′(c) = 5, specifically c = 1 or c = 3/2.
To verify that the Intermediate Value Theorem for Derivatives applies to the function f′(x) on the interval [0,2], we need to check if f′(x) is continuous on the interval [0,2].
Since f′(x) is the derivative of f(x), and f(x) is differentiable everywhere, it implies that f′(x) is continuous on [0,2].
Now, we need to find the value of c guaranteed by the theorem such that f′(c) = 5.
To do this, we can use the concept of the Mean Value Theorem. According to the Mean Value Theorem, if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) such that f′(c) = (f(b) - f(a))/(b - a).
In our case, we have f′(x) = 4x^2 - 4x + 2 and we want to find a value c such that f′(c) = 5.
Applying the Mean Value Theorem, we have:
(f(2) - f(0))/(2 - 0) = f′(c)
Now, let's calculate f(2) and f(0).
f(2) = ∫[0, 2] f′(x) dx
= ∫[0, 2] (4x^2 - 4x + 2) dx
= [4/3 x^3 - 2x^2 + 2x] from 0 to 2
= (4/3 * 2^3 - 2 * 2^2 + 2 * 2) - (0)
= (4/3 * 8 - 2 * 4 + 4)
= (32/3 - 8 + 4)
= 32/3 - 4/3
= 28/3
f(0) = ∫[0, 0] f′(x) dx
= 0
Substituting these values, we have:
(28/3 - 0)/(2 - 0) = f′(c)
(28/3)/(2) = 5
(28/3)(1/2) = 5
28/6 = 5
Simplifying further:
14/3 = 5
Since there is no value c that satisfies this equation, there is no value of c in the interval [0,2] such that f′(c) = 5.
Thus, the Intermediate Value Theorem for Derivatives does not apply to the function f′(x) on the interval [0,2] in this case.
To verify that the Intermediate Value Theorem for Derivatives applies to the function f′(x) on the interval [0, 2], we need to check if f′(x) is continuous on that interval.
To find the value of c guaranteed by the theorem such that f′(c) = 5, we can use the fact that f′(x) is continuous and apply the Intermediate Value Theorem for Derivatives.
Steps to find the value of c:
1. First, we need to check if f′(x) is continuous on the interval [0, 2].
- To do this, we need to check if f′(x) is defined and continuous at every point in the interval [0, 2]. Since f′(x) = 4x^2 - 4x + 2 is a polynomial function, it is defined and continuous everywhere. So, f′(x) is continuous on the interval [0, 2].
2. Now, we can apply the Intermediate Value Theorem for Derivatives.
- The Intermediate Value Theorem for Derivatives states that if a function f′(x) is continuous on the closed interval [a, b] and if the values of f′(a) and f′(b) have opposite signs, then there exists a value c in the interval (a, b) such that f′(c) = 0.
- Since we are looking for a value c such that f′(c) = 5, we need to find two points in the interval [0, 2] where f′(x) takes on values with opposite signs.
3. Plug in x = 0 and x = 2 into f′(x) to find their corresponding values.
- When x = 0: f′(0) = 4(0)^2 - 4(0) + 2 = 2
- When x = 2: f′(2) = 4(2)^2 - 4(2) + 2 = 14
4. Since f′(0) = 2 and f′(2) = 14 have opposite signs (one positive and one negative), there exists a value c in the interval (0, 2) such that f′(c) = 5.
To find the exact value of c, we can use an iterative process or numerical methods such as Newton's Method to approximate the value of c. However, this cannot be determined without further information or calculations.