Problem 58 - Algebra and Trigonometry

Question

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. $$ g(x)=x^{5}-8 x^{3}+20 x $$

Step-by-Step Solution

Answer

The local maximum is 11.31 at \( x = \sqrt{2} \) and the local minimum is -11.31 at \( x = -\sqrt{2} \).

Step 1: Find the First Derivative

To find the local maximum and minimum values of the function, we first need to find the derivative of the function. The function is given as \( g(x) = x^5 - 8x^3 + 20x \). Using basic differentiation rules, we find the first derivative: \[ g'(x) = 5x^4 - 24x^2 + 20 \]

Step 2: Set the First Derivative to Zero

To find the critical points, set the first derivative equal to zero and solve for \( x \):\[ 5x^4 - 24x^2 + 20 = 0 \]This is a quartic equation, but can be treated as a quadratic in terms of \( x^2 \), let \( u = x^2 \). So the equation becomes:\[ 5u^2 - 24u + 20 = 0 \]

Step 3: Solve the Quadratic Equation

The quadratic equation \( 5u^2 - 24u + 20 = 0 \) can be solved using the quadratic formula:\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 5, b = -24, c = 20 \). Calculating the discriminant and solving, we find:\[ u = 2 \quad \text{and} \quad u = 2 \]Thus, \( x^2 = 2 \). This gives us the solutions \( x = \sqrt{2} \) and \( x = -\sqrt{2} \).

Step 4: Evaluate the Second Derivative

To identify the nature of the critical points, compute the second derivative:\[ g''(x) = 20x^3 - 48x \]

Step 5: Check the Concavity at Critical Points

Plug the critical points \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) into the second derivative to determine the concavity:- For \( x = \sqrt{2} \), \( g''(\sqrt{2}) = 20(\sqrt{2})^3 - 48(\sqrt{2}) < 0 \), indicating a local maximum.- For \( x = -\sqrt{2} \), \( g''(-\sqrt{2}) = 20(-\sqrt{2})^3 - 48(-\sqrt{2}) > 0 \), indicating a local minimum.

Step 6: Calculate Function Values

Find the function values at the critical points:- For \( x = \sqrt{2} \), \( g(\sqrt{2}) = (\sqrt{2})^5 - 8(\sqrt{2})^3 + 20(\sqrt{2}) \).- For \( x = -\sqrt{2} \), \( g(-\sqrt{2}) = (-\sqrt{2})^5 - 8(-\sqrt{2})^3 + 20(-\sqrt{2}) \).Performing these calculations:- At \( x = \sqrt{2} \), \( g(\sqrt{2}) \approx 11.31 \).- At \( x = -\sqrt{2} \), \( g(-\sqrt{2}) \approx -11.31 \).

Key Concepts

Local Maxima and Minima

In calculus, understanding the concepts of local maxima and minima is essential in determining the behavior of a function over a specific interval. A local maximum is a point where a function's value is higher than all surrounding points in the immediate area. Conversely, a local minimum is where the function's value is lower than the surrounding points in its immediate vicinity. These are crucial because they reveal the peaks and troughs in a graph, helping to understand underlying trends or cycles in any applied context. For the function given, we find these points by first determining where the slope of the tangent is zero, then evaluating the concavity to decide if these points are peaks or troughs.

First Derivative

The first derivative gives us a way to measure the rate of change of a function at any point along its curve. It is calculated by differentiating the function with respect to a variable, typically denoted as \( g'(x) \). The result tells us how steep the slope is at any given point.

• When the first derivative is positive, the function is increasing at that point.
• When it is negative, the function is decreasing.
• Where it equals zero, we might find a local maximum or minimum, but not always.

For instance, the derivative of the function \( g(x) = x^5 - 8x^3 + 20x \) is \( g'(x) = 5x^4 - 24x^2 + 20 \). Solving \( g'(x) = 0 \) will help us locate the critical points, which are the potential candidates for local maxima or minima.

Critical Points

Critical points are identified where the first derivative of a function equals zero or where the derivative does not exist. These points suggest where the function might experience horizontal tangents, leading to local extremum.

• The location of these points is crucial for further analysis.
• They are found by solving the equation \( g'(x) = 0 \).

In our example, solving \( 5x^4 - 24x^2 + 20 = 0 \) gives \( x^2 = 2 \), leading to the critical points \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). Each represents a spot where the function may change direction: transitioning from increasing to decreasing, or vice versa.

Second Derivative Test

The second derivative test helps clarify the nature of the critical points we find. By using the second derivative \( g''(x) \), we can understand the concavity of the function at any critical point.

• If \( g''(x) > 0 \), the function is concave up, indicating a local minimum.
• If \( g''(x) < 0 \), the function is concave down, indicating a local maximum.
• If \( g''(x) = 0 \), the test is inconclusive.

For this problem, we evaluated the second derivative for \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) using \( g''(x) = 20x^3 - 48x \). We found that \( x = \sqrt{2} \) led to a negative value, suggesting it's a local maximum, and \( x = -\sqrt{2} \) gave a positive outcome, indicating a local minimum.