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.
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 .
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