Problem 34 - Essential Calculus

Question

Find all relative extrema. Use the Second Derivative Test where applicable. \(g(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-3)^{2} / 2}\)

Step-by-Step Solution

Answer

The function \(g(x)\) has a local minimum at \(x = 3\).

Step 1: Find the first derivative

First, find the derivative of \(g(x)\). It's noteworthy that the Chain Rule will be used here. The derivative \(g'(x)\) can be found via using chain rule: \(g'(x) = -\frac{(x-3)}{\sqrt{2\pi}}e^{-(x-3)^2 / 2}\).

Step 2: Locate potential extrema

Set the derivative equal to zero and solve for x to find critical points. Here, \(g'(x) = 0\) gives \(x-3 = 0\), which in turn gives \(x = 3\). So, this is a potential extremum.

Step 3: Find the second derivative

Continue finding the second derivative, \(g''(x)\). Use the product rule and the chain rule again: \(g''(x) = \frac{1}{\sqrt{2\pi}}e^{-(x-3)^2 / 2} - \frac{(x-3)^2}{\sqrt{2\pi}}e^{-(x-3)^2 / 2}\)

Step 4: Confirm extrema using second derivative test

The second derivative test states that if \(g''(c) > 0\), then \(g(c)\) is a local minimum, and if \(g''(c) < 0\), then \(g(c)\) is local maximum. The value of the second derivative at \(x = 3\) is positive, so \(x = 3\) is a local minimum point.

Key Concepts

Relative Extrema

When we talk about relative extrema, we refer to the highest or lowest points within a particular region of a function. These points are crucial because they represent places where the function changes direction from increasing to decreasing or vice versa.

Understanding relative extrema involves:

• **Finding critical points:** These are points where the function's derivative equals zero or is undefined.
• **Using the second derivative test:** This is used to determine the nature (maximum or minimum) of the critical points found.

For example, if at a point, the first derivative is zero, and the second derivative is positive, this indicates a relative minimum. If the second derivative is negative, we have a relative maximum.
In the context of the problem with the function \( g(x) \), finding extrema involved setting the first derivative to zero, yielding the critical point at \( x = 3 \). Further, using the second derivative test revealed that \( x = 3 \) is a relative minimum.

Chain Rule

The Chain Rule is a fundamental technique in calculus utilized to find the derivative of a composed function. If you have a function that is the result of another function being applied, the Chain Rule helps break it down into simpler derivatives.
The Chain Rule states:

• If \( y = f(g(x)) \), then the derivative \( y' \) is given by \( f'(g(x)) \times g'(x) \).

In simpler terms, first derive the outer function with respect to the inner function, and then multiply it by the derivative of the inner function.
In the solution to the exercise with \( g(x) \), the Chain Rule helped us derive the Gaussian function, where the function \( e^{-(x-3)^2 / 2} \) was differentiated by treating \( -(x-3)^2 / 2 \) as the inner function.

Critical Points

Critical points are essential in identifying where a function's graph may have peaks, valleys, or flat spots. A critical point occurs where the first derivative of a function equals zero or is undefined, indicating a potential change in the direction of the graph.
To find critical points:

• Calculate the first derivative of the function.
• Solve for \( x \) where the derivative equals zero or does not exist.

For the function in question, \( g'(x) \) was set to zero which led us to find \( x = 3 \). This was identified as our potential extremum or critical point.
Assessing critical points is a preliminary step before using further tests, like the second derivative test, to confirm the nature of the extrema.

Product Rule

The Product Rule is another vital concept in calculus used when differentiating products of two functions. When two functions are multiplied together, their derivative can be found using this rule.
The Product Rule states:

• If \( h(x) = f(x)g(x) \), then the derivative \( h'(x) \) is \( f'(x)g(x) + f(x)g'(x) \).

This means we take the derivative of the first function and multiply it by the second, add this to the first function multiplied by the derivative of the second function.
In our problem, the Product Rule was applied while finding \( g''(x) \), the second derivative, to accommodate both parts of the expression. Handling these compounded expressions effectively helped attain the accurate mathematical form required for testing the extrema at their critical point.