Problem 25 - Essential Calculus Early Transcendentals

Question

Does there exist a function \(f\) such that \(f(0)=-1\) \(f(2)=4,\) and \(f^{\prime}(x) \leqslant 2\) for all \(x ?\)

Step-by-Step Solution

Answer

No, such a function cannot exist because the required slope exceeds the derivative constraint.

Step 1: Understand the Problem

We are looking for a continuous function \( f \) that starts at \( f(0) = -1 \), ends at \( f(2) = 4 \), and has a derivative \( f'(x) \leq 2 \) for all \( x \). The function's maximum slope is constrained by the derivative condition.

Step 2: Calculate the Required Average Slope

Calculate the slope needed to connect \( f(0) = -1 \) and \( f(2) = 4 \). The slope (or average rate of change) is given by: \[ m = \frac{f(2) - f(0)}{2 - 0} = \frac{4 - (-1)}{2} = \frac{5}{2} = 2.5 \] . This slope must be achieved at some point if \( f \) is linear.

Step 3: Evaluate the Derivative Condition

The derivative of any calculus function describes its slope. According to the problem, \( f'(x) \leq 2 \). Compare this with the required slope 2.5 from step 2. Since 2.5 > 2, \( f \) cannot be formed under this derivative constraint.

Step 4: Conclusion About the Existence of \( f \)

Since the required average slope of 2.5 exceeds the maximum allowable slope of 2 given by \( f'(x) \leq 2 \), no such function \( f \) can exist.

Key Concepts

Average Rate of Change

The concept of the average rate of change provides a simple way to measure how a function's output changes as the input changes over a specified interval. Imagine plotting a line between two points on a curve; this line is known as the secant line.

• To compute the average rate of change from point \( a \) to \( b \), use the formula: \( \frac{f(b) - f(a)}{b-a} \).
• In our example, with two points given by \( f(0) = -1 \) and \( f(2) = 4 \), the average rate of change is \( \frac{4 - (-1)}{2 - 0} = 2.5 \).
• This value, 2.5, tells us how much the function changes per unit interval over \([0, 2]\).

This average rate of change is critical because it sets a benchmark for what the function's slope must achieve to bridge the two points.

Derivative

In calculus, the derivative of a function at a particular point quantifies the instantaneous rate of change of the function at that point. It's like zooming in to see the slope of the tangent line to the curve.

• The derivative is not just an average like the rate of change; rather, it provides the exact slope of the function at any given point.
• In this problem, the constraint \( f'(x) \leq 2 \) establishes a cap on how steeply the function can rise. This means wherever you look on the graph of \( f \), the slope must not exceed 2.
• However, the calculated average slope of 2.5 from the previous section shows the function would need to change faster than the derivative allows.

Understanding the derivative constraint helps assess whether a function can attain the required average rate of change under its maximum allowable rate.

Slope of a Function

The slope of a function is an essential idea in both algebra and calculus. It determines how a function increases or decreases at a particular point or over an interval.

• In geometry, the slope of a line is defined as \( m = \frac{\text{rise}}{\text{run}} \), representing the steepness and direction of the line.
• In calculus, the slope becomes a function of \( x \), known as \( f'(x) \), providing a dynamic aspect as this slope can change from one point to another on the curve.
• While a constant function has a uniform slope, many functions do not. They can have regions where they rise quickly and others where they level out or even descend.

For the given exercise, the slope over the interval \([0, 2]\) needed to be a consistent 2.5 for a linear function; however, the restriction \( f'(x) \leq 2 \) prevents achieving the required rise efficiently within that interval.