Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ f(x)=x^{5}-\frac{25}{2} x^{3}+20 x-1 ; I=[-3,2] $$

Short answer

The maximum value is at \( x = 2 \) and the minimum at \( x = -3 \).

Step-by-step solution

Find the derivative

First, find the derivative of the function \( f(x) = x^5 - \frac{25}{2}x^3 + 20x - 1 \). The derivative, \( f'(x) \), is calculated as follows: \[ f'(x) = 5x^4 - \frac{75}{2}x^2 + 20 \].

Set derivative to zero and solve

Set the derivative equal to zero to find the critical points: \[ 5x^4 - \frac{75}{2}x^2 + 20 = 0 \]. Solve this equation for \( x \). It can be simplified by dividing all terms by 5, producing: \[ x^4 - \frac{15}{2}x^2 + 4 = 0 \]. Let \( y = x^2 \), then you have a quadratic equation in terms of \( y \): \[ y^2 - \frac{15}{2}y + 4 = 0 \]. Solve for \( y \) using the quadratic formula: \[ y = \frac{15 \pm \sqrt{\left( \frac{15}{2} \right)^2 - 4 \cdot 1 \cdot 4}}{2} \].

Find the solutions for y

Continue solving for \( y \): \[ y = \frac{15 \pm \sqrt{\left( \frac{15}{2} \right)^2 - 16}}{2} \]. Simplify: \[ y = \frac{15 \pm \sqrt{56.25 - 16}}{2} \], which further simplifies to \[ y = \frac{15 \pm \sqrt{40.25}}{2} \]. Calculate \( \sqrt{40.25} \), approximate it as \( 6.35 \), leading to \[ y_1 = \frac{15 + 6.35}{2}, \quad y_2 = \frac{15 - 6.35}{2} \]. So, \( y_1 \approx 10.675 \) and \( y_2 \approx 4.325 \).

Translate y back to x

Recall that \( y = x^2 \), so you need to find \( x \) for \( y_1 \approx 10.675 \) and \( y_2 \approx 4.325 \). For \( y_1 \approx 10.675 \), \( x_1 = \pm \sqrt{10.675} \). For \( y_2 \approx 4.325 \), \( x_2 = \pm \sqrt{4.325} \). Approximating, \( \sqrt{10.675} \approx 3.27 \) and \( \sqrt{4.325} \approx 2.08 \). Hence, the potential critical points are \( x_1 \approx 3.27, -3.27, 2.08, -2.08 \).

Check interval validity

The interval given is \( I = [-3, 2] \). The derived \( x \)-values need to be within this interval. The valid values are \( x = -2.08, 2 \) from the calculated approximate critical points that lie within \([-3, 2]\). Include the endpoints of the interval for evaluation: \( x = -3, 2 \).

Evaluate the function at critical points and endpoints

Evaluate the original function \( f(x) \) at the critical points within the interval and at the endpoints: - \( f(-3) = (-3)^5 - \frac{25}{2}(-3)^3 + 20(-3) - 1 \)- \( f(-2.08) = f(-2.08) \) (compute for the value)- \( f(2) = 2^5 - \frac{25}{2}(2)^3 + 20(2) - 1 \)Calculate these values to find the minimum and maximum values.

Determine maximum and minimum values

Compare the function values calculated at the critical points and endpoints within the interval to identify the maximum and minimum values. This includes checking \( f(-3), f(-2.08), f(2) \), and any determined valid critical values.

Key concepts

Derivative of a Function

The derivative of a function, often denoted as \( f'(x) \), helps us understand how a function is changing at any given point. It is the foundation for finding critical points where the function may achieve maximum or minimum values.

To calculate the derivative, differentiate each term of the function separately. For example, if we have a function \( f(x) = x^5 - \frac{25}{2}x^3 + 20x - 1 \), take the derivative of each component:

• The derivative of \( x^5 \) is \( 5x^4 \).
• The derivative of \( -\frac{25}{2}x^3 \) is \( -\frac{75}{2}x^2 \).
• The derivative of \( 20x \) is \( 20 \).
• The derivative of the constant \( -1 \) is \( 0 \).

Putting these together, the derivative \( f'(x) \) is \( 5x^4 - \frac{75}{2}x^2 + 20 \).

Finding the derivative is a crucial step because it allows us to solve for critical points where a function's slope (or rate of change) equals zero, indicating where the graph may have a horizontal tangent line.

Finding Maximum and Minimum Values

After finding the derivative, set it equal to zero and solve for \( x \) to find the critical points. For example, if you have \( f'(x) = 5x^4 - \frac{75}{2}x^2 + 20 \), set the equation:

\[ 5x^4 - \frac{75}{2}x^2 + 20 = 0 \]

Next, simplify and solve the equation. Here, we simplify by dividing through by 5:

\[ x^4 - \frac{15}{2}x^2 + 4 = 0 \]

This can be transformed to a simpler quadratic form by letting \( y = x^2 \), resulting in:

\[ y^2 - \frac{15}{2}y + 4 = 0 \]

Use the quadratic formula to solve for \( y \), and then backtrack to find \( x \). This gives you potential critical points. Evaluating these points and the function at endpoints of the given interval helps determine the maximum and minimum values. Critical points that lie within the interval are analyzed to see if they offer maximum or minimum values by plugging them back into the original function and comparing outputs.

Intervals in Calculus

Intervals in calculus are specific sections of the number line defined by two endpoints, such as \( [-3, 2] \). When analyzing functions, it's important to determine which critical points fall within these intervals.

Our task is to evaluate these points within the interval by plugging them into the original function \( f(x) \), especially the potential critical points found by solving \( f'(x) = 0 \). Equally important are the endpoints because the maximum or minimum can also occur there.

For example, consider an interval \( I = [-3, 2] \), you check:

• The function's behavior at potential critical points like \( -2.08 \) and points such as \( -3 \) and \( 2 \).

By evaluating the original function at these specific \( x \) values, you determine where it reaches its highest or lowest within the specified range, aiding in the accurate determination of relative extrema. This ensures you don't miss out on critical evaluations for potential maximum or minimum values.