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Suppose that \(f^{\prime}(x) \leq 1\) for \(1 \leq x \leq 4 .\) Show that \(f(4)-\) \(f(1) \leq 3 .\)

Short Answer

Expert verified
Given the derivative restriction and MVT, \(f(4) - f(1) \leq 3.\)

Step by step solution

01

Understand the Exercise

The exercise provides that the derivative of a function, \(f'(x),\) is less than or equal to 1 for the interval \(1 \leq x \leq 4.\) We need to show that the difference \(f(4) - f(1)\) does not exceed 3.
02

Recall the Mean Value Theorem

The Mean Value Theorem (MVT) 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: \(f'(c) = \frac{f(b) - f(a)}{b - a}.\)
03

Apply the Mean Value Theorem

For our function \(f(x)\), \(a = 1\) and \(b = 4\). Apply the MVT: \(f'(c) = \frac{f(4) - f(1)}{4 - 1} = \frac{f(4) - f(1)}{3}.\)
04

Use the Given Inequality

We're informed that \(f'(x) \leq 1\) for all \(x\) in \([1, 4].\) Therefore, \(f'(c) \leq 1\) for some \(c\) in \((1, 4),\) implying that \(\frac{f(4) - f(1)}{3} \leq 1.\)
05

Solve the Inequality

Multiply both sides of the inequality \(\frac{f(4) - f(1)}{3} \leq 1\) by 3 to isolate \(f(4) - f(1):\)\[f(4) - f(1) \leq 3.\]
06

Conclusion

We have demonstrated that \(f(4) - f(1) \leq 3\), using the Mean Value Theorem and given conditions about the derivative. This verifies the statement in the exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
The concept of a derivative is foundational in calculus, representing the rate of change of a function with respect to its variable. When you see a notation like \( f'(x) \), it indicates the derivative of \( f(x) \) at any point \( x \). This value tells you how quickly \( f(x) \) changes when \( x \) changes slightly. In the context of the exercise, the derivative \( f'(x) \leq 1 \) suggests that the function \( f(x) \) does not increase faster than a slope of 1 across the specified interval \([1, 4]\). This is important because it directly limits how much the function can change over a distance. The derivative being bounded by 1 ensures that the function's growth is controlled, implying that large unexpected spikes or drops in the function are not possible in this section of its graph.
Inequality
Inequalities are mathematical statements indicating that one expression is less than or greater than another. They play a crucial role in bounding or limiting values. In this exercise, the inequality \( f'(x) \leq 1 \) helps restrict the potential change in the function \( f(x) \). Using the Mean Value Theorem, which provides a bridge between derivatives and the actual change in function values, we apply this inequality. When the theorem uses \( f'(c) = \frac{f(4) - f(1)}{3} \) to represent the average rate of change over \([1, 4]\), the inequality \( f'(c) \leq 1 \) guarantees \( \frac{f(4) - f(1)}{3} \leq 1 \).Applying this inequality leads us directly to our main result: solving it by multiplying both sides by 3, yielding \( f(4) - f(1) \leq 3 \). Through these steps, inequalities help manage the range of possible values \( f(x) \) can take.
Function Continuity
Function continuity is a fundamental property in calculus, meaning there are no gaps, jumps, or sudden breaks in the graph of a function \( f(x) \). A function is continuous over an interval if you can draw it from start to finish without lifting your pen. The significance of continuity in this exercise comes from the Mean Value Theorem, which requires \( f(x) \) to be continuous over \([1, 4]\) and differentiable over \((1, 4)\). This ensures that the function behaves predictably; there are no unexpected breaks where differentiability could not be applied.In the context of our exercise, continuity implies that the function smoothly transitions from \( f(1) \) to \( f(4) \). This smoothness is part of why the Mean Value Theorem can confidently predict the existence of a point \( c \) where the derivative represents the overall rate of change across the entire interval.

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