In the realm of mathematical analysis, understanding the behavior of functions and their derivatives is pivotal. This article delves into an intriguing problem involving a continuous function f defined on the interval [1, 4], which is also differentiable on (1, 4). The function satisfies specific conditions: f(1) = 1, f(2) = 5, f(3) = 9, and f(4) = 17. Additionally, we are given a crucial constraint: |f'(x) - 4| + |f(x) - 4| ≠ 0 for all x in the open interval (1, 4). Our primary objective is to determine the minimum number of roots of the equation f'(x) = 0.
This exploration combines concepts from calculus, such as continuity, differentiability, and the Mean Value Theorem, with logical reasoning to deduce the properties of f'(x). The condition |f'(x) - 4| + |f(x) - 4| ≠ 0 plays a significant role, as it provides a constraint on the relationship between the function's derivative and its value. This article will provide a comprehensive analysis and solution to unveil the minimum number of roots for f'(x) = 0 under these specific conditions.
Let f: [1, 4] → ℝ be a continuous function on [1, 4] and differentiable on (1, 4) with f(1) = 1, f(2) = 5, f(3) = 9, and f(4) = 17. Given that |f'(x) - 4| + |f(x) - 4| ≠ 0 for all x ∈ (1, 4), determine the minimum number of roots of the equation f'(x) = 0.
Applying the Mean Value Theorem
To begin, let's leverage the Mean Value Theorem (MVT) on different subintervals of [1, 4]. The MVT states that if a function is continuous on [a, b] and differentiable on (a, b), there exists at least one point c in (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
We will apply this theorem to the intervals [1, 2], [2, 3], and [3, 4].
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Interval [1, 2]:
Applying the MVT on [1, 2], there exists a c₁ ∈ (1, 2) such that:
f'(c₁) = (f(2) - f(1)) / (2 - 1) = (5 - 1) / 1 = 4
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Interval [2, 3]:
Applying the MVT on [2, 3], there exists a c₂ ∈ (2, 3) such that:
f'(c₂) = (f(3) - f(2)) / (3 - 2) = (9 - 5) / 1 = 4
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Interval [3, 4]:
Applying the MVT on [3, 4], there exists a c₃ ∈ (3, 4) such that:
f'(c₃) = (f(4) - f(3)) / (4 - 3) = (17 - 9) / 1 = 8
We have now established that there exist points c₁ ∈ (1, 2), c₂ ∈ (2, 3), and c₃ ∈ (3, 4) where f'(c₁) = 4, f'(c₂) = 4, and f'(c₃) = 8.
Utilizing the Given Condition
We are given the condition |f'(x) - 4| + |f(x) - 4| ≠ 0 for all x ∈ (1, 4). This implies that for any x in (1, 4), f'(x) cannot be 4 if f(x) is also 4, and vice versa. In other words, f'(x) and f(x) cannot both be equal to 4 at the same x. This condition provides a crucial constraint on the behavior of the function and its derivative.
Analyzing the Function Values
Let's analyze the values of f(x) at the given points: f(1) = 1, f(2) = 5, f(3) = 9, and f(4) = 17. Since f is continuous on [1, 4], by the Intermediate Value Theorem, for any value between 1 and 17, there must exist at least one point in the interval [1, 4] where f(x) takes that value. Specifically, there must exist points x in (1, 4) such that f(x) = 4.
To confirm the existence of points where f(x) = 4, we can examine the intervals:
- On the interval [1, 2], f(1) = 1 and f(2) = 5. By the Intermediate Value Theorem, there exists at least one point x₁ ∈ (1, 2) where f(x₁) = 4.
- On the interval [2, 3], f(2) = 5 and f(3) = 9, so it is possible that f(x) > 4 on this entire interval. However, we know there exists a point x₁ ∈ (1,2) where f(x₁) = 4.
- On the interval [3, 4], f(3) = 9 and f(4) = 17, so f(x) > 4 on this interval.
Since there exists an x₁ ∈ (1, 2) where f(x₁) = 4, the condition |f'(x₁) - 4| + |f(x₁) - 4| ≠ 0 implies that |f'(x₁) - 4| ≠ 0, so f'(x₁) ≠ 4. This is an important deduction.
Rolle's Theorem and Existence of Roots
Now, let’s consider the derivatives we found using the Mean Value Theorem:
- f'(c₁) = 4 for some c₁ ∈ (1, 2)
- f'(c₂) = 4 for some c₂ ∈ (2, 3)
- f'(c₃) = 8 for some c₃ ∈ (3, 4)
We know c₁ ∈ (1, 2) and there exists x₁ ∈ (1, 2) such that f(x₁) = 4. Since f'(c₁) = 4 and f'(x₁) ≠ 4, we have two distinct points in the interval (1, 2) where the derivative takes different values. This suggests the derivative must change sign within this interval, indicating the existence of at least one root where f'(x) = 0.
Specifically, let’s analyze the interval between c₁ and x₁. If c₁ < x₁, we have f'(c₁) = 4 and f'(x₁) ≠ 4. If f'(x₁) > 4, then by the Intermediate Value Theorem applied to the continuous function f'(x), there must be a point between c₁ and x₁ where f'(x) takes the value 0. If f'(x₁) < 4, then considering the condition f'(x) ≠ 4 when f(x) = 4, there must be a point where f'(x) = 0. A similar argument applies if x₁ < c₁.
Next, consider the interval (2, 3). We have f'(c₂) = 4 and f'(c₃) = 8 for c₂ ∈ (2, 3) and c₃ ∈ (3, 4). To ensure f'(x) becomes 0, we need to analyze the possible changes in f'(x) between c₂ and c₃.
Since f'(c₂) = 4 and f'(c₃) = 8, we can consider the function g(x) = f(x) - 4x. We have g'(x) = f'(x) - 4. The given condition |f'(x) - 4| + |f(x) - 4| ≠ 0 tells us that at any point where f'(x) = 4, f(x) ≠ 4. This helps us infer behavior but does not directly guarantee a root. However, combined with MVT, we see
- f(2) = 5, so g(2) = f(2) - 4*2 = 5 - 8 = -3
- f(3) = 9, so g(3) = f(3) - 4*3 = 9 - 12 = -3
By Rolle's Theorem applied to g(x) on [2, 3], since g(2) = g(3) = -3, there exists a c in (2, 3) such that g'(c) = 0. Therefore, f'(c) - 4 = 0, which implies f'(c) = 4.
Now consider the interval [c₂, c₃] where c₂ ∈ (2, 3) and c₃ ∈ (3, 4). We have f'(c₂) = 4 and f'(c₃) = 8. Since f'(x) is continuous (as f is differentiable), by the Intermediate Value Theorem, there must be a point in (c₂, c₃) where f'(x) takes every value between 4 and 8. To reach 0, f'(x) must decrease at some point. So, there must be at least one point in (c₂, c₃) where f''(x) is negative.
Considering our initial deduction from c₁ and x₁, we identified that f'(x) must equal 0 at least once in (1, 2). Also, to transition from 4 to 8, there needs to be an inflection point where f''(x) changes sign. This implies that f'(x) must decrease from 4 and increase again to 8. This change suggests that f'(x) has to reach 0 at least one more time.
Conclusion
Given these deductions, we can conclude that the minimum number of roots of the equation f'(x) = 0 is 1 in the interval (1,2) due to the condition and Intermediate Value Theorem applied to the derivative, and at least another one to change from 4 to 8. Therefore, there must be at least two points where f'(x) = 0. Thus, the minimum number of roots for f'(x) is 1.
Based on the analysis conducted, we have determined that the minimum number of roots of the equation f'(x) = 0, under the given conditions, is 1. This conclusion is derived from applying the Mean Value Theorem, Intermediate Value Theorem, and the provided constraint |f'(x) - 4| + |f(x) - 4| ≠ 0. The interplay between these concepts leads to a comprehensive understanding of the behavior of f'(x) and its roots within the specified interval.
- The Mean Value Theorem was applied to the subintervals [1, 2], [2, 3], and [3, 4] to find points where f'(x) takes specific values, such as 4 and 8.
- The condition |f'(x) - 4| + |f(x) - 4| ≠ 0 ensures that f'(x) and f(x) cannot both be 4 at the same point, influencing the possible behavior of f'(x).
- The Intermediate Value Theorem helped establish the existence of points where f(x) = 4, which, in conjunction with the given condition, implies the existence of roots for f'(x).
- By analyzing the changes in f'(x) and applying logical deductions, we determined that f'(x) must equal 0 at least once in the interval (1, 4).
- We proved that the minimum number of roots for f'(x) = 0 is 1.
This detailed analysis demonstrates how a combination of calculus principles and logical reasoning can solve complex problems involving functions and their derivatives. The insights gained from this problem can be applied to similar scenarios in mathematical analysis and related fields.