Problem 42 - Calculus

Question

(a) Graph \(f(x)=\frac{1}{2} x^{2}\) and \(g(x)=f(x)+3\) on the same set of axes. What can you say about the slopes of the tangent lines to the two graphs at the point \(x=0 ? x=2 ?\) Any point \(x=x_{0} ?\) (b) Explain why adding a constant value, \(C,\) to any function does not change the value of the slope of its graph at any point. [Hint: Let \(g(x)=f(x)+C\) and calculate the difference quotients for \(f \text { and } g .]\)

Step-by-Step Solution

Answer

The slopes of both functions at any point are equal, as adding a constant doesn't change the derivative.

Step 1: Graph the Functions

First, we need to graph the functions. The function \( f(x) = \frac{1}{2}x^2 \) is a parabola that opens upwards with its vertex at the origin (0,0). The second function \( g(x) = f(x) + 3 = \frac{1}{2}x^2 + 3 \) is the same parabolic curve as \( f(x) \), but it is shifted upwards by 3 units.To graph these functions:1. Plot the key points of \( f(x) \), such as (-2, 2), (-1, 0.5), (0, 0), (1, 0.5), and (2, 2).2. For \( g(x) \), add 3 to the y-coordinates of these points, resulting in (-2, 5), (-1, 3.5), (0, 3), (1, 3.5), and (2, 5). 3. Draw smooth curves through these points to create the parabolic shapes.

Step 2: Analyze the Slopes at Specific Points

To analyze the slopes of the tangent lines, we need to find the derivatives of the functions. The derivative of \( f(x) = \frac{1}{2}x^2 \) is \( f'(x) = x \). The derivative of \( g(x) = \frac{1}{2}x^2 + 3 \) is \( g'(x) = x \), since the derivative of a constant is zero.Now calculate the slopes at given points:- At \( x = 0 \), the slope \( f'(0) = 0 \) and \( g'(0) = 0 \). Thus, the slopes are the same.- At \( x = 2 \), the slope \( f'(2) = 2 \) and \( g'(2) = 2 \). Again, the slopes are the same.- At any point \( x = x_0 \), the slope \( f'(x_0) = x_0 \) and \( g'(x_0) = x_0 \). So, the slopes are equal.

Step 3: Justification When Adding a Constant

To prove that adding a constant does not change the slope, consider the difference quotients:For \( f(x) \), the difference quotient is \( \frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{2}(x+h)^2 - \frac{1}{2}x^2}{h} \).For \( g(x) \), which is \( g(x)=f(x)+C \), the difference quotient is \( \frac{g(x+h) - g(x)}{h} = \frac{\left( \frac{1}{2}(x+h)^2 + C \right) - \left( \frac{1}{2}x^2 + C \right)}{h} = \frac{\frac{1}{2}(x+h)^2 - \frac{1}{2}x^2}{h} \).Notice that the constants cancel out, thus both difference quotients are the same. Therefore, adding a constant does not affect the derivative.

Key Concepts

Graphing Functions

Graphing functions is a fundamental skill in calculus that allows you to visualize mathematical expressions. In this example, we have two functions to graph:

• \(f(x) = \frac{1}{2}x^2\)
• \(g(x) = f(x) + 3 = \frac{1}{2}x^2 + 3\)

Both functions represent parabolas that open upwards. The first function, \(f(x)\), has its vertex at the origin (0,0). The second function, \(g(x)\), is simply shifted upwards by 3 units because of the constant addition. To graph both: plot a series of points based on their equations.

For \(f(x)\), choose key points like (-2, 2), (-1, 0.5), (0, 0), (1, 0.5), and (2, 2). For \(g(x)\), add 3 to the y-coordinates of these points to get (-2, 5), (-1, 3.5), (0, 3), (1, 3.5), and (2, 5). With these points, draw a smooth curve to form the parabolic shapes.

This shifting of the graph upwards is typical when adding a constant to a function. Understanding these shifts is essential as it forms the basis for interpreting more complex transformations.

Derivatives

Derivatives are a central concept in calculus and are used to determine the rate of change of a function. They can tell you how quickly a curve is rising or falling at any given point.

In our exercise, we find the derivatives of two functions:

• \(f(x) = \frac{1}{2}x^2\)
• \(g(x) = \frac{1}{2}x^2 + 3\)

Both derivatives are the same: \(f'(x) = x\) and \(g'(x) = x\). This is because the derivative of a constant, like 3 in \(g(x)\), is zero.

The derivative \(f'(x) = x\) for these functions tells us that the slope of the tangent to the curve at any point \(x\) is \(x\). What this means is at \(x = 0\), the rate of change is 0, indicating no slope, while at \(x = 2\), the rate of change, or slope, is 2, indicating an upward slope.

Derivatives provide crucial insights into function behavior and allow us to understand the nature of changing systems.

Tangent Lines

Tangent lines are straight lines that touch a curve at a single point without crossing it. They represent the instant rate of change at that specific point.

In this exercise, we examine tangent lines for functions \(f(x)\) and \(g(x)\) at different points. The derivative acts as the slope of the tangent line:

• At \(x = 0\), the slopes for both functions are 0, so the tangent is a horizontal line.
• At \(x = 2\), the slope is 2 for both functions, indicating a tangent that slopes upwards.
• At any general point \(x = x_0\), the slope is \(x_0\), showing the consistency across the original and shifted function.

These tangent lines are crucial in understanding instantaneous rates, such as velocity in physics or growth in economics. Knowing how to compute and interpret these lines assists in practical and theoretical analysis of dynamic systems.

Constant Addition

Adding a constant to a function shifts its graph vertically without altering its shape or slope at any point. Let's explore why this is the case.

By adding a constant \(C\) to a function, \(g(x) = f(x) + C\), we shift its entire graph up or down by \(C\) units. However, what remains unchanged is the derivative because the rate of change of the function itself doesn't depend on the constant.

When we compute the difference quotient for both \(f(x)\) and \(g(x) = f(x) + C\), we observe that the constant \(C\) cancels out in the calculation. This results in the identical derivative: all the slope information relies on the base function \(f(x)\).

• This is why, graphically, the slope (and therefore the tangent) at any specific point \(x\) remains unchanged even after adding a constant.
• Such concepts are vital in understanding how alterations in equations affect their visual and mathematical properties.

Understanding the impact of constant additions helps in broadening perspective on transformations and reinforces the notion of slope being purely dependent on the non-constant portions of functions.