Problem 24 - Essential Calculus Early Transcendentals

Question

Suppose that 3\(\leqslant f^{\prime}(x) \leqslant 5\) for all values of \(x .\) Show that \(18 \leqslant f(8)-f(2) \leqslant 30\)

Step-by-Step Solution

Answer

\(18 \leq f(8)-f(2) \leq 30\) is confirmed with MVT.

Step 1: Understanding the Derivative Bounds

We know that the derivative of a function, denoted as \( f'(x) \), gives the rate of change of the function. The problem states that the derivative \( f'(x) \) is bounded between 3 and 5 for all \( x \). This means:\[3 \leq f^{\prime}(x) \leq 5\]This implies that for any change in \( x \), the change in \( f(x) \) will be between 3 times and 5 times that of the change in \( x \).

Step 2: Applying the Mean Value Theorem (MVT)

The Mean Value Theorem tells us for a function \( f \) that is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists some \( c \in (a, b) \) such that:\[f'(c) = \frac{f(b) - f(a)}{b-a}\]In this problem, let \( a = 2 \) and \( b = 8 \). By the Mean Value Theorem:\[3 \leq \frac{f(8) - f(2)}{8 - 2} \leq 5\]which simplifies to:\[3 \leq \frac{f(8) - f(2)}{6} \leq 5\]

Step 3: Solving the Inequality

Multiply all parts of the inequality by 6 to solve for \( f(8) - f(2) \):\[3 \times 6 \leq f(8) - f(2) \leq 5 \times 6\]This simplifies to:\[18 \leq f(8) - f(2) \leq 30\]This matches the given condition.

Key Concepts

Derivative Bounds

The concept of derivative bounds is essential in understanding how a function changes. When we talk about the derivative of a function, denoted as \( f'(x) \), we're essentially discussing the function's rate of change at any given point.

• If we know that \( 3 \leq f'(x) \leq 5 \) for all \( x \), it tells us that the function's rate of change is never slower than 3 or faster than 5.
• Think of this as the function having a "speed limit." The change in \( f(x) \) as \( x \) increases is somewhere between 3 times and 5 times the change in \( x \).

This is crucial because it sets boundaries for the behavior of the function, letting us make predictions about the function's output based on how much \( x \) changes. With this bound, we ensure that no sharp or unexpected spikes occur in the graph of the function.

Rate of Change

Understanding the rate of change involves applying the Mean Value Theorem (MVT), which is a cornerstone of calculus.

• The MVT states that if a function \( f \) is continuous on the interval \([a, b]\) and differentiable on \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) where the instantaneous rate of change \( f'(c) \) equals the average rate of change over that interval.
• This can be written as \( f'(c) = \frac{f(b) - f(a)}{b-a} \).

Applying this to our problem, where \( a = 2 \) and \( b = 8 \), means finding the average rate of change from \( x = 2 \) to \( x = 8 \) and ensuring it's within the bounds we discussed earlier (3 and 5). Thus, \( 3 \leq \frac{f(8)-f(2)}{6} \leq5 \). The MVT essentially ensures us that the average speed of the function between \( a \) and \( b \) aligns with the derivative bounds.

Inequality

Solving inequalities in calculus is akin to solving a puzzle where the pieces are interconnected by mathematical logic. Once we have established our inequality from the Mean Value Theorem, we need to manipulate it to directly find the range within which a specific change in the function must lie.

• In our case, starting with \( 3 \leq \frac{f(8) - f(2)}{6} \leq 5 \), our task is to solve for \( f(8) - f(2) \).
• To do this, we multiply all parts of the inequality by the same number (6, in this case) to isolate \( f(8) - f(2) \). This operation yields \( 18 \leq f(8) - f(2) \leq 30 \).

This final inequality gives us a range that shows how much \( f(8) \) is related to \( f(2) \) after applying the derivative bounds and the Mean Value Theorem. It assures us that the change from \( x = 2 \) to \( x = 8 \) in the function is neither too small nor too large, adhering to our initially determined rate of change limits.