Question

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

Short answer

Maximum value is 19; minimum value is -1.

Step-by-step solution

Find the Derivative

To find the critical points, first find the derivative of the function. For the function \( f(x) = x^3 - 3x + 1 \), the derivative is \( f'(x) = 3x^2 - 3 \).

Set Derivative to Zero

Set the derivative \( f'(x) = 3x^2 - 3 \) equal to zero to find critical points: \( 3x^2 - 3 = 0 \). Divide both sides by 3 to simplify: \( x^2 - 1 = 0 \).

Solve for Critical Points

Solve the equation \( x^2 - 1 = 0 \) by factoring: \( (x - 1)(x + 1) = 0 \). The solutions are \( x = 1 \) and \( x = -1 \).

Check Critical Points in the Interval

Check if the critical points \( x = 1 \) and \( x = -1 \) are within the given interval \( \left(-\frac{3}{2}, 3\right) \). Both points are in the interval.

Evaluate the Function at Critical Points and Endpoints

Evaluate \( f(x) \) at critical points and endpoints:- \( f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^3 - 3\left(-\frac{3}{2}\right) + 1 = \frac{-27}{8} + \frac{9}{2} + 1 = \frac{23}{8} \).- \( f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 \).- \( f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1 \).- \( f(3) = (3)^3 - 3(3) + 1 = 27 - 9 + 1 = 19 \).

Identify Maximum and Minimum Values

Compare the values obtained:- Maximum value is 19 at \( x = 3 \).- Minimum value is -1 at \( x = 1 \).

Key concepts

Derivative Calculation

To identify critical points of a function, the first step is to calculate its derivative. The derivative tells us how the function's output changes relative to changes in its input. Understanding this change is crucial for finding critical points, which are where the function's slope is zero or undefined. For the function given,

• Start with the function: \( f(x) = x^3 - 3x + 1 \).
• Apply the power rule to find the derivative: \( f'(x) = 3x^2 - 3 \).

This expression represents the rate of change of the function. Setting the derivative to zero will reveal the function's critical points, key for understanding where the function may have peaks or troughs.

Maximum and Minimum Values

Once the critical points are determined, the next task is to identify where the function attains its maximum and minimum values. These values help us understand the range of the function's behavior and are essential in various optimization problems.

• Critical Points: Found by solving \( f'(x) = 0 \).
• Endpoints: Consider the values of the function at the edges of the interval, here being \( x = -\frac{3}{2} \) and \( x = 3 \).

Evaluate the function at each of these points to compare values. The highest function value among these points will be the maximum value, and the lowest function value will be the minimum value on the given interval.

Interval Analysis

Interval analysis in calculus involves examining a function within a specified range to understand its behavior. Evaluating functions over predetermined intervals can provide insights into possible maximum and minimum values within that range.

• First, find the critical points using the derivative of the function.
• Second, ensure that these critical points lie within the given interval: \( \left(-\frac{3}{2}, 3\right) \).

For our function, both critical points, \( x = 1 \) and \( x = -1 \), lie within the interval, which makes them relevant for finding maximum and minimum values. Any endpoint values within this interval should also be checked.

Function Evaluation

After identifying the critical points and ensuring they are within the interval, the next phase is to evaluate the function at these points and any endpoints. Function evaluation is key to determining the actual maximum and minimum values.

• Calculate \( f(x) \) at each critical point and each endpoint: \( f(-1), f(1), f\left(-\frac{3}{2}\right), f(3) \).
• This step involves substituting the points back into the original function.

By comparing these results, we find the maximum of 19 at \( x = 3 \) and the minimum of -1 at \( x = 1 \), thereby characterizing the function's behavior over the interval.

Factoring Equations

Factoring is a method used to simplify equations, especially useful when setting derivatives to zero to find critical points. In the context of our problem, the derivative equation was factored to ease finding the roots.

• Start with the simplified equation: \( 3x^2 - 3 = 0 \).
• Divide through by 3 to simplify further: \( x^2 - 1 = 0 \).
• Factor this as: \((x - 1)(x + 1) = 0 \).

This factoring reveals the solutions \( x = 1 \) and \( x = -1 \), which are crucial as they pinpoint where the slope of the function is zero, indicating potential local maxima or minima within the analyzed interval.