Problem 72 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the relative maxima and relative minima, if any, of each function. $$ g(x)=\frac{x}{x^{2}-1} $$

Step-by-Step Solution

Answer

The function \(g(x) = \frac{x}{x^2 - 1}\) does not have any relative maxima or relative minima as there are no critical points within the domain of the function.

Step 1: Find the first derivative

To find the first derivative, we will use the quotient rule which states that if \(h(x) = \frac{f(x)}{g(x)}\), then: \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\) Applying the quotient rule to our function, \(g(x) =\frac{x}{x^2 - 1}\), we have: \(g'(x) = \frac{(1)(x^2-1) - (x)(2x)}{(x^2-1)^2}\)

Step 2: Determine the critical points

Now, we set the first derivative to 0 and solve for x: \(0 = \frac{(1)(x^2-1) - (x)(2x)}{(x^2-1)^2}\) To solve for \(x\), we will only focus on the numerator since setting denominator equal to zero does not provide critical points: \(0 = x^2 -1 - 2x^2 \) Now, simplify the equation and solve for \(x\): \(0 = -x^2 - 1\) \(x^2 = 1\) Critical points are \(x = 1\) and \(x = -1\), but note these points are not in the domain of the original function, so we have no critical points in this case.

Step 3: Identify additional critical points

Check for points at which the first derivative is not differentiable. In our case, the derivative g'(x) is not differentiable at x = 1 and x = -1, which are the points where the denominator becomes zero. However, we already excluded these points from the domain of the original function. Since we have no critical points in the domain of the given function, we can conclude that the function does not have any relative maxima or relative minima.

Key Concepts

quotient rule

The quotient rule is a handy technique in calculus used to find the derivative of a function that is expressed as a quotient of two other functions, say \(h(x) = \frac{f(x)}{g(x)}\). When tackling calculus problems like these, understanding how to apply the quotient rule is crucial. It allows you to manage complex fractions easily.
To apply the quotient rule, remember the formula:

• Find \(f'(x)\) and \(g'(x)\), which are the derivatives of the top and bottom functions, respectively.
• Substitute these into the derivative formula: \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\).

In our example, for the function \(g(x) = \frac{x}{x^2-1}\), we identified:

• \(f(x) = x\), thus \(f'(x) = 1\).
• \(g(x) = x^2-1\), thus \(g'(x) = 2x\).

By applying these into the quotient rule formula, we found the derivative \(g'(x)\) as a tidy step towards further analyzing the function.

critical points

Critical points in calculus are points on the graph of a function where the derivative is either zero or undefined. These points are significant because they can signal a change in the behavior of the function, typically associated with local peaks, troughs, or points of inflection.
To determine these points:

• Set the first derivative to zero and solve for \(x\).
• Identify where the derivative is undefined.

In our example function \(g(x) = \frac{x}{x^2-1}\), after applying the quotient rule, set \(g'(x) = 0\). Simplifying this, we arrived at potential critical points being \(x = \pm 1\).
However, these points are not valid within the domain of the original function since they would make the denominator zero, causing the function to be undefined. Thus, no critical points exist within the function's domain, meaning we won't find potential maxima or minima from this route.

relative maxima and minima

Relative maxima and minima provide insight into where a function reaches its highest or lowest points relative to the surrounding values. These are the peaks and valleys one can observe on a graph, pivotal for understanding the function's overall behavior.
Typically, relative maxima or minima occur at critical points, where the derivative is zero or undefined.
To find these points, check:

• That the points are within the function's domain.
• If the derivative changes sign around these critical points (use the first derivative test).

In our case with \(g(x) = \frac{x}{x^2-1}\), while \(x = 1\) and \(x = -1\) were candidates, they lie outside the domain, as they make the function undefined. Since our critical points don't exist within the function's domain, there are no relative maxima or minima. Usually, if such points existed in the domain, they could indicate points where the function's value starts increasing or decreasing significantly around those \(x\)-values.