Problem 57 - Algebra and Trigonometry

Question

\(51-58=\) Find the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. $$ V(x)=\frac{1-x^{2}}{x^{3}} $$

Step-by-Step Solution

Answer

Local minima occur at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) with values \( -0.39 \) and \( 0.39 \), respectively.

Step 1: Find the Derivative

To find the local maxima and minima, we first need to find the derivative of the function. Given \( V(x) = \frac{1-x^{2}}{x^{3}} \), apply the quotient rule. The quotient rule is \( \left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^{2}} \). Here, \( u = 1-x^2 \) and \( v = x^3 \). Thus, \( u' = -2x \) and \( v' = 3x^2 \). The derivative is: \[ V'(x) = \frac{x^3 (-2x) - (1-x^2)(3x^2)}{x^6} \] Simplify this to get: \[ V'(x) = \frac{-2x^4 - 3x^2 + 3x^4}{x^6} = \frac{x^4 - 3x^2}{x^6} = \frac{x^2 (x^2 - 3)}{x^6} = \frac{x^2 - 3}{x^4} \].

Step 2: Solve for Critical Points

Set the derivative \( V'(x) = 0 \) to find critical points:\[ \frac{x^2 - 3}{x^4} = 0 \]This simplifies to \( x^2 - 3 = 0 \), which means:\[ x^2 = 3 \]Solving for \( x \), we get \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).

Step 3: Test for Local Maxima and Minima

Use the second derivative test to identify local maxima or minima. First, find the second derivative \( V''(x) \). Differentiate \( V'(x) = \frac{x^2 - 3}{x^4} \).\[ V''(x) = \left( \frac{(2x)(x^4) - (x^2 - 3)(4x^3)}{x^8} \right)\]Simplifying gives:\[ V''(x) = \frac{2x^5 - 4x^5 + 12x^3}{x^8} = \frac{-2x^5 + 12x^3}{x^8} = \frac{-2x^2 + 12}{x^6} \].Calculate \( V''(x) \) at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \):For \( x = \sqrt{3} \): \[ V''(\sqrt{3}) = \frac{-6 + 12}{9} = \frac{6}{9} = \frac{2}{3} > 0 \], indicating a local minimum.For \( x = -\sqrt{3} \): \[ V''(-\sqrt{3}) = \frac{-6 + 12}{9} = \frac{6}{9} = \frac{2}{3} > 0 \], indicating a local minimum also.

Step 4: Calculate the Function Values

Substitute the critical points back into the original function to find their values:For \( x = \sqrt{3} \):\[ V(\sqrt{3}) = \frac{1-(\sqrt{3})^2}{(\sqrt{3})^3} = \frac{1-3}{3\sqrt{3}} = \frac{-2}{3\sqrt{3}} \approx -0.385 \].For \( x = -\sqrt{3} \):\[ V(-\sqrt{3}) = \frac{1-(-\sqrt{3})^2}{(-\sqrt{3})^3} = \frac{1-3}{-3\sqrt{3}} = \frac{-2}{-3\sqrt{3}} = \frac{2}{3\sqrt{3}} \approx 0.385 \].

Key Concepts

Quotient Rule

The quotient rule is an essential concept in calculus for differentiating functions that are expressed as a ratio of two distinct functions. If you have a function written as \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions, then the derivative of the function \( \left( \frac{u}{v} \right)' \) is found using the following formula:

• \( \left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^2} \)

To apply the quotient rule, you must:

• Identify \( u \) and \( v \) from your function. In our example, \( u = 1 - x^2 \) and \( v = x^3 \).
• Compute the derivatives \( u' \) and \( v' \). In the example, \( u' = -2x \) and \( v' = 3x^2 \).
• Plug these values into the quotient rule formula to find \( V'(x) \).

After putting everything together, you can simplify to find the derivative of the original function. It's a useful rule for complex functions where the numerator and denominator are not simply constants.

Derivatives

Derivatives give you the rate at which a function is changing at any given point, essentially providing a mathematical way to capture 'instant speed' of change. When working with a function like \( V(x) = \frac{1-x^2}{x^3} \), finding the derivative helps in determining critical points where local maxima or minima can occur.
To determine the derivative \( V'(x) \), the approach involves breaking down the function by identifying distinct components, such as using the chain, product, or quotient rule. In our case, the quotient rule was utilized.
The practical aspect of finding derivatives lies in its application: once you have \( V'(x) \), you set it to zero to solve for critical points, where the function changes its rate of increase or decrease. These critical points are essential when determining the behavior of the function across its domain, helping to identify where the function peaks or valleys.

Second Derivative Test

The second derivative test is a mathematical method used to classify critical points determined from the first derivative. Once the first derivative \( V'(x) \) is found and set to zero to determine critical points, taking the second derivative \( V''(x) \) helps identify whether these points are peaks, troughs, or saddle points.
The process involves finding \( V''(x) \) by differentiating \( V'(x) \) again. In this case, the calculation was made from \( V'(x) = \frac{x^2 - 3}{x^4} \), leading to \( V''(x) = \frac{-2x^2 + 12}{x^6} \).
Once \( V''(x) \) is found, the sign of the second derivative needs to be evaluated at each critical point:

• If \( V''(x) > 0 \), there's a local minimum at that point.
• If \( V''(x) < 0 \), there's a local maximum.

In our example, both critical points \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) exhibited positive second derivatives, indicating local minima at these positions.

Local Maximum and Minimum

Local maxima and minima are significant points where a function reaches a peak or valley, but not necessarily the extreme overall. They help understand the behavior of the function within a certain range.
In this context, after finding the critical points using \( V'(x) \), and classifying them using the second derivative test, you substitute these points back into the original function to find their respective function values. This step is critical to comprehending not just where the peaks and valleys are, but their magnitude.
To illustrate, substituting \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) into \( V(x) = \frac{1-x^2}{x^3} \) revealed values of local minima. Specifically:

• For \( x = \sqrt{3} \), \( V(\sqrt{3}) \approx -0.385 \)
• For \( x = -\sqrt{3} \), \( V(-\sqrt{3}) \approx 0.385 \)

Understanding local maxima and minima is crucial for applications ranging from economic models to physics, wherever determining optimal points within a real range is required.