Problem 79 - Calculus Early Transcendentals

Question

Functions similar to $$ f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} $$ arise in a wide variety of statistical problems. (a) Use the first derivative test to show that \(f\) has a relative maximum at \(x=0,\) and confirm this by using a graphing utility to graph \(f\) (b) Sketch the graph of $$ f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2} $$ where \(\mu\) is a constant, and label the coordinates of the relative extrema.

Step-by-Step Solution

Answer

The function has a relative maximum at \(x = 0\). The translated function's peak is at \(x = \mu\).

Step 1: Find the First Derivative

To find the first derivative of the function \(f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\), use the chain rule. The derivative of \(e^{-x^2/2}\) with respect to \(x\) is \(-x e^{-x^2/2}\). Thus, the derivative is: \[f'(x) = \frac{1}{\sqrt{2\pi}} (-x) e^{-x^2/2} = -\frac{x}{\sqrt{2\pi}} e^{-x^2/2}.\]

Step 2: Set the Derivative to Zero

To find potential extrema, set the derivative \(f'(x)\) to zero: \[-\frac{x}{\sqrt{2\pi}} e^{-x^2/2} = 0.\] Since \(e^{-x^2/2} eq 0\) for all \(x\) (as the exponential function is always positive), the only way for the equation to be zero is if \(x = 0\).

Step 3: First Derivative Test

To use the first derivative test, check the sign of \(f'(x)\) on either side of \(x = 0\): - For \(x < 0\), \(-\frac{x}{\sqrt{2\pi}} > 0\), so \(f'(x) > 0\). - For \(x > 0\), \(-\frac{x}{\sqrt{2\pi}} < 0\), so \(f'(x) < 0\). Since \(f'(x)\) changes from positive to negative at \(x = 0\), there is a relative maximum at \(x = 0\).

Step 4: Graph Verification

Using a graphing utility to plot \(f(x)\), verify that there is a peak at \(x = 0\). The function's bell-shaped structure reflects this observation, confirming a relative maximum at \(x = 0\).

Step 5: Sketch the Graph of Modified Function

For the modified function \(f(x) = \frac{1}{\sqrt{2\pi}} e^{-(x-\mu)^2/2}\), recognize it is a horizontal translation of the original graph by \(\mu\) units. The relative maximum is now at \(x = \mu\).

Step 6: Label the Relative Extrema

On the sketch of the translated graph \(f(x)\), label the coordinates of the relative maximum (\(\mu, \frac{1}{\sqrt{2\pi}}\)). The point reflects the new position of the peak after the horizontal shift.

Key Concepts

First Derivative Test

The first derivative test is a technique used in calculus to determine where a function may have relative extrema, such as relative minima or maxima. To apply the first derivative test, you first need to find the derivative of a function and look for the points at which this derivative equals zero or does not exist. These are your critical points.

For the function in our exercise, the first derivative was found using the chain rule, resulting in:

• \( f'(x) = -\frac{x}{\sqrt{2\pi}} e^{-x^2/2} \)

Next, set the first derivative equal to zero to find critical points:

• \( -\frac{x}{\sqrt{2\pi}} e^{-x^2/2} = 0 \)
• \( \rightarrow x = 0 \)

Now, use the first derivative test to determine if this critical point is a max or a min. Check the sign of the derivative before and after the critical point:

• For \( x < 0 \), \( f'(x) > 0 \) — the function is increasing
• For \( x > 0 \), \( f'(x) < 0 \) — the function is decreasing

Because the derivative changes from positive to negative as it passes through \( x = 0 \), this indicates a relative maximum at \( x = 0 \). This behavior is graphically confirmed, showing a peak at this point.

Exponential Functions

Exponential functions are central in many mathematical contexts due to their unique properties. The general form of an exponential function is \( f(x) = a \cdot e^{bx} \) where \( e \) is the base of natural logarithms, approximately equal to 2.718.

The function in our exercise:

• \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \)

is particularly notable because of its role in the Gaussian distribution, a cornerstone in statistics.

Some key characteristics of exponential functions include:

• They never touch the x-axis, as \( e^x \) is always positive.
• They exhibit continuous growth or decay, as determined by the sign of the exponent.
• In our modified function, \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-(x-\mu)^2/2} \), we see a transformation that shifts the peak according to \( \mu \), demonstrating a horizontal shift.

These features make exponential functions versatile in modeling phenomena such as population growth, radioactive decay, and, as seen here, probability distributions.

Graphing Functions

Graphing functions is a crucial skill in understanding the behavior of mathematical models visually. By plotting the function \( f(x) \), you can gain insights into its dynamics, such as intervals of increase and decrease, maxima, and minima.

For our function:

• \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \)

Using a graphing utility displays a bell-shaped curve with a clear peak at \( x = 0 \). This visual representation confirms the mathematical analysis provided by the first derivative test. As the curve approaches its peak, it rises rapidly, indicating an increase in the function's value before quickly decreasing after reaching the maximum.

When graphing the modified function:

• \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-(x-\mu)^2/2} \)

The graph shifts horizontally to align its peak with \( x = \mu \). This transformation emphasizes the effect of translating the graph along the x-axis, reinforcing how simple changes in the equation can significantly impact its graph.

Effective graphing involves:

• Identifying critical points and asymptotes.
• Understanding transformations such as translations and reflections.
• Using technology to enhance comprehension of complex functions.

These components contribute to a comprehensive understanding of mathematical behavior through graphical representation.