Problem 166
Question
Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(x) \geq 2\) for \(x \in[1,6]\), then \(\quad\) [2005] (A) \(f(6) \geq 8\) (B) \(f(6)<8\) (C) \(f(6)<5\) (D) \(f(6)=5\)
Step-by-Step Solution
Verified Answer
(A) \( f(6) \geq 8 \) is correct.
1Step 1: Understand the Problem
We are given that the function \( f \) is differentiable for all \( x \). It is also known that \( f(1) = -2 \) and the derivative \( f'(x) \geq 2 \) for \( x \in [1,6] \). We need to determine the correct statement about \( f(6) \).
2Step 2: Apply the Mean Value Theorem
The Mean Value Theorem states that for a differentiable function \( f \) on the interval \([a,b]\), there exists \( c \in (a,b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b-a} \). Here, we apply it on the interval \([1,6]\).
3Step 3: Set Up the Inequality
Using the Mean Value Theorem, we have:\[f'(c) = \frac{f(6) - f(1)}{6-1} = \frac{f(6) + 2}{5}.\]Since it is given that \( f'(x) \geq 2 \), we have\[\frac{f(6) + 2}{5} \geq 2.\]
4Step 4: Solve the Inequality
Multiply both sides by 5 to clear the fraction:\[f(6) + 2 \geq 10.\]Subtract 2 from both sides to solve for \( f(6) \):\[f(6) \geq 8.\]
5Step 5: Conclusion
Based on our solution, \( f(6) \geq 8 \). Therefore, the correct answer is option (A): \( f(6) \geq 8 \).
Key Concepts
Mean Value TheoremInequalityDerivative concept
Mean Value Theorem
The Mean Value Theorem is a key result in calculus, particularly when discussing differentiability and the behavior of functions over an interval. This theorem essentially says that for any given section of a continuous and differentiable curve, there is at least one point where the instantaneous rate of change (the derivative) matches the average rate of change over that interval. This important concept is often encapsulated in the formula:
- If a function \( f \) is continuous over \([a, b]\) and differentiable over \((a, b)\), then there persists a point \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b-a} \).
Inequality
Inequalities are expressions that show the relative size or order of two values. When dealing with inequalities in calculus, they often express constraints or conditions that a function or its derivative must satisfy. In the given exercise, the statement \( f'(x) \geq 2 \) for \( x \in [1,6] \) represents an inequality constraint on the derivative of the function.
- An inequality like \( f'(x) \geq 2 \) suggests that the rate of change of the function is never less than 2 within the interval.
- This can affect the behavior of the function dramatically, ensuring that \( f(x) \) is increasing at a rate of at least 2 units per unit increase in \( x \).
Derivative concept
Understanding derivatives is central to calculus and mathematical analysis. The derivative represents the rate at which a quantity changes. It's often described as the slope of the tangent line at any point on a graph.
- A derivative, denoted as \( f'(x) \), provides us with a way to determine the behavior of the function—whether it's increasing, decreasing, or constant.
- In the context of the exercise, \( f'(x) \geq 2 \) implies a steady or accelerating increase in the function within the interval \([1,6]\).
Other exercises in this chapter
Problem 164
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A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched? Interval Function (A
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A spherical iron ball \(10 \mathrm{~cm}\) in radius is coated with a layer of ice of uniform thickness than melts at a rate of \(50 \mathrm{~cm}^{3} / \mathrm{m
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The function \(f(x)=\frac{x}{2}+\frac{2}{x}\) has a local minimum at (A) \(x=2\) (B) \(x=-2\) (C) \(x=0\) (D) \(x=1\)
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