Problem 1 - Calculus

Question

(a) Estimate \(f^{\prime}(2)\) using the values of \(f\) in the table. (b) For what values of \(x\) does \(f^{\prime}(x)\) appear to be positive? Negative? $$\begin{array}{c|c|c|c|c|c|c|c} \hline x & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline f(x) & 10 & 18 & 24 & 21 & 20 & 18 & 15 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

(a) \(f'(2) \approx 3.5.\) (b) \(f'(x)>0\) for \(0 < x < 4\); \(f'(x)<0\) for \(4 < x < 12\).

Step 1: Understand Derivative Estimate

To estimate the derivative of a function at a certain point, you can use the average rate of change between two nearby points. This is given by the formula: \(f'(x) \approx \frac{f(b) - f(a)}{b - a}\). It estimates how fast \(f\) is changing at \(x=a\).

Step 2: Choose Points for Estimation at \(x = 2\)

To estimate \(f'(2)\), choose \(x = 0\) and \(x = 4\) as nearby points since these lie on either side of 2 in the table. The function values are \(f(0) = 10\), \(f(2) = 18\), and \(f(4) = 24\).

Step 3: Calculate \(f'(2)\)

Using the points \((0, 10)\) and \((4, 24)\), compute the slope: \[f'(2) \approx \frac{f(4) - f(0)}{4 - 0} = \frac{24 - 10}{4} = \frac{14}{4} = 3.5.\] This value, 3.5, estimates the derivative at \(x = 2\).

Step 4: Analyze \(f(x)\) for Increasing and Decreasing Intervals

\(f\) is increasing where the function values are rising as \(x\) increases and decreasing where function values are falling. From the table: \(f\) increases from \(x = 0\) to \(x = 4\) (values: 10 to 24) and decreases from \(x = 4\) onwards.

Step 5: Identify where \(f'(x)\) is Positive

\(f'(x)\) is positive where \(f\) is increasing. In the given table, between \(x = 0\) and \(x = 4\), \(f(x)\) goes from 10 to 24, which implies \(f'(x) > 0\).

Step 6: Identify where \(f'(x)\) is Negative

\(f'(x)\) is negative where \(f\) is decreasing. From \(x = 4\) to \(x = 12\), \(f(x)\) decreases from 24 down to 15, implying \(f'(x) < 0\).

Key Concepts

Understanding Average Rate of Change

The average rate of change of a function represents the way the function's output values change as its input values change. It's like measuring the speed of a car by comparing how far it travels over a period of time. In mathematical terms, given a function \(f(x)\), the average rate of change between two points \(a\) and \(b\) is given by the formula: \[ f'(x) \approx \frac{f(b) - f(a)}{b - a} \] This formula essentially finds the slope of the line connecting the function's values at points \(a\) and \(b\). The closer these points are to each other, the more accurate the estimate for the derivative at a particular point, \(x\), becomes.

• If the average rate of change is positive, the function is generally increasing in that interval.
• If it is negative, the function is generally decreasing.

In the given exercise, to estimate \(f'(2)\), we analyze points around \(x = 2\), specifically \(x = 0\) and \(x = 4\). This gives us an average rate of change, or slope, of 3.5.

Analyzing Increasing and Decreasing Functions

Functions can be described as increasing or decreasing based on how their outputs change as we move along the x-axis. If the function values rise as we move from left to right, the function is increasing in that particular region. Conversely, if the values fall, the function is decreasing. Looking at the table in the exercise:

• The function \(f(x)\) increases from \(x = 0\) to \(x = 4\), which means the values go up from 10 to 24.
• Once we move past \(x = 4\), the function starts to decrease, as the values reduce from 24 down to 15 up to \(x = 12\).

Identifying these intervals assists in understanding where the derivative of the function (or \(f'(x)\)) will be positive or negative. When a function is increasing, the derivative is positive; When it's decreasing, the derivative is negative.

Positive and Negative Derivatives

Derivatives are a way of indicating how quickly a function is changing at any given point. The sign of the derivative—whether it's positive or negative—tells us the direction of this change.

• Positive derivatives: These signify that the function is increasing. In our exercise, this occurs between \(x = 0\) and \(x = 4\), as the function value rises from 10 to 24. This is why \(f'(x)\) is positive in this interval.
• Negative derivatives: These indicate that the function is decreasing. According to the table, from \(x = 4\) to \(x = 12\), the function drops from 24 to 15, implying \(f'(x)\) is negative.

Understanding when a derivative is positive or negative helps in predicting the behavior of the function and its graph. Recognizing this change is crucial in applications involving motion, growth, and other real-world situations.