Problem 14 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

If we have a formula for \(f\) we can get quite good numerical estimates of the slope of the tangent line to \(f\) at a particular point. In this exercise we will do that. Below is the graph of \(y=f(x)=x^{2}-4\) (a) Goal: We want to estimate the value of \(f^{\prime}(2)\), i.e., to approximate the slope of the tangent line to the graph at the point \((2,0)\). To this end, do the following: (b) Goal: We want to estimate the value of \(f^{\prime}(1) ;\) that is, we want to approximate the slope of the tangent line to the graph at the point \((1,-3)\). (c) Goal: We want to estimate the value of \(f^{\prime}(0) ;\) i.e., we want to approximate the slope of the tangent line to the graph at the point \((0,-4)\). i. Sketch the tangent line to the graph at \((0,-4)\). ii. Find the slope of the line through \((0,-4)\) and \((0+h, f(0+h))\). iii. Use your answer to part ii to estimate the slope of the tangent line. (d) Goal: To estimate the value of \(f^{\prime}(\mathrm{c})\) for \(\mathrm{c}\) a constant. i. Find the slope of the line through (c, \(f(\mathrm{c}))\) and \((\mathrm{c}+h, f(\mathrm{c}+h))\). ii. Use your answer to the previous question to nd the slope of the tangent line to \(f\) at \(x=\mathrm{c}\). iii. Sketch the graph of \(f^{\prime}(x)\).

Step-by-Step Solution

Answer

\(\left. f'(2) = 2, f'(1) = 2, f'(0) = 0, \: \text{and} \: f'(c) = 2c \right.\)

Step 1: Find \(f’(2)\)

We will compute the difference quotient: \(f'_{h}(x) = \frac{{f(x+h)-f(x)}}{h}\). We have \(f(2 + h) - f(2) = (2 + h)^{2} - 4 -(2^{2}-4) = 2h + h^{2}\). So when \(x = 2\), then \(f'_{h}(2) = \frac{{2h+h^{2}}}{h} = 2 + h\). Therefore, as \(h\) approaches 0, it follows that \(f'(2) = 2\).

Step 2: Find \(f’(1)\)

Similarly as step 1, we calculate the difference quotient: \(f'_{h}(x) = \frac{{f(x+h)-f(x)}}{h}\). We have \(f(1 + h) - f(1) = (1 + h)^{2} - 4 -(1^{2}-4) = 2h + h^{2}\). So when \(x = 1\), then \(f'_{h}(1) = \frac{{2h+h^{2}}}{h} = 2 + h\). So, as \(h\) approaches 0, we get \(f'(1) = 2\).

Step 3: Find \(f’(0)\)

We calculate the difference quotient: \(f'_{h}(x) = \frac{{f(x+h)-f(x)}}{h}\). We have \(f(0 + h) - f(0) = (0 + h)^{2} - 4 -(0^{2}-4) = h^{2} + 4\). So when \(x = 0\), then \(f'_{h}(0) = \frac{{h^{2}+4}}{h} = h + 4\). So, as \(h\) approaches 0, we find \(f'(0) = 0\).

Step 4: Find \(f’(c)\)

By generalizing the earlier work, we can find that when \(x = c\), using the same structure as the previous steps, the difference quotient would be \(f'_{h}(c) = \frac{{f(c+h)-f(c)}}{h}\) = \(2c + h\). As \(h\) approaches 0, we find \(f'(c) = 2c\).

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus, providing an essential step towards understanding rates of change. It represents the average rate of change of a function over an interval, as the interval approaches zero.

The difference quotient is given by the formula: \[ \frac{f(x + h) - f(x)}{h} \] where \( f(x) \) is the function in question, \( x \) is a point on the domain of \( f \), and \( h \) is an increment in \( x \). It can be seen as the slope of the secant line connecting two points on the function's graph: \((x, f(x))\) and \((x+h, f(x+h))\).

As \( h \) gets smaller and approaches zero, the secant line gets closer to the tangent line at point \( x \), and the difference quotient approaches the derivative. This process reflects the underlying idea of a limit in calculus. Demonstrating this with an example from the exercise, for the function \(f(x) = x^2 - 4\), a difference quotient for \( f'(2) \) would be \[ f'_h(2) = \frac{{f(2+h) - f(2)}}{h} \]

Derivative Calculation

The derivative of a function at a point is the limit of the difference quotient as \( h \) approaches zero. It provides an exact rate of change or slope of the tangent line at that point. Calculating the derivative involves finding this limit, if it exists.

In the context of the exercise given, we can demonstrate this with the function \(f(x) = x^2 - 4\). For instance, to calculate \(f'(2)\), we determine the limit of the difference quotient as follows: \[ \lim_{h \to 0} f'_h(2) = \lim_{h \to 0} \frac{{(2 + h)^{2} - 4 - (2^{2} - 4)}}{h} \] Simplification leads to \(f'(2) = 2\). This result signifies the slope of the tangent line of \(f(x)\) at \(x = 2\).

Derivative calculations are a cornerstone in calculus, allowing us to understand how functions change and to solve problems involving motion, optimization, and more.

Slope of Tangent Line

The slope of the tangent line to the graph of a function at a given point is essentially the derivative of the function at that point. It represents how steep the line is, and it is a specific, instantaneous rate of change.

Using the derivative of \(f(x) = x^2 - 4\), which we have calculated as \(f'(x) = 2x\), we can find the slope of the tangent line at any point \(c\) by simply plugging in the value of \(c\). For example, the slope of the tangent line at \(x = 1\) is \(f'(1) = 2(1) = 2\).

The visualization of this concept is also critical. For \(x = 0\), sketching the graph reveals that the slope of the tangent line is horizontal, confirming that \(f'(0) = 0\), as found in the step-by-step solution. Learning how to assess and sketch the behavior of the slope of the tangent line can greatly enhance a student's intuition about the function's behavior.