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

The minimum value is \(-1\) and the maximum value is \(19\).

Step-by-step solution

Find the derivative of the function

To find the critical points, first, we need to find the derivative of the given function. The function is \( f(x) = x^3 - 3x + 1 \). By applying the power rule, \( f'(x) = 3x^2 - 3 \).

Set the derivative to zero

To find the critical points, set the derivative equal to zero: \( 3x^2 - 3 = 0 \). Simplify and solve for \( x \) by factoring: \( 3(x^2 - 1) = 0 \). This simplifies to \( 3(x - 1)(x + 1) = 0 \). Thus, \( x = 1 \) and \( x = -1 \) are the critical points.

Evaluate the function at critical points and endpoints

Now calculate \( f(x) \) at the critical points and the endpoints of the interval \([-\frac{3}{2}, 3]\). Evaluate:- \( f(-\frac{3}{2}) = (-\frac{3}{2})^3 - 3(-\frac{3}{2}) + 1 = -\frac{27}{8} + \frac{9}{2} + 1 = \frac{11}{8} \).- \( f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 \).- \( f(1) = 1^3 - 3(1) + 1 = 1 - 3 + 1 = -1 \).- \( f(3) = 3^3 - 3(3) + 1 = 27 - 9 + 1 = 19 \).

Identify the maximum and minimum values

To find the maximum and minimum values, compare the values found in Step 3. The function values are: \( f(-\frac{3}{2}) = \frac{11}{8} \), \( f(-1) = 3 \), \( f(1) = -1 \), and \( f(3) = 19 \). Therefore, the minimum value is \( f(1) = -1 \) and the maximum value is \( f(3) = 19 \).

Key concepts

Interval Notation

Interval notation is a way to describe a set of numbers along a number line. It is used to define the domain or range of a function, especially when we talk about critical points within a given segment. In our exercise, the interval is specified as \( I = \left[-\frac{3}{2}, 3\right] \).
This means:

• The interval starts at \(-\frac{3}{2}\) and ends at \(3\).
• It's a closed interval since both fix values have brackets \([]\) around them, indicating that the endpoints are included.

This notation helps to clearly identify where we are interested in finding the critical points and evaluating the function. By including endpoints, we ensure that any potential maximum or minimum values at these points are considered.

Function Derivative

The derivative of a function is a crucial concept in calculus. It represents the rate at which the function's value changes as the input changes. For the function \( f(x) = x^3 - 3x + 1 \), we need to find its derivative \( f'(x) \) to identify the critical points.
To find the derivative:

• Apply the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \).
• So, for \( x^3 \), the derivative is \( 3x^2 \).
• For \(-3x\), the derivative is \(-3\).
• The constant \(1\) becomes \(0\), as constants have no rate of change.

Thus, the derivative of our function is \( f'(x) = 3x^2 - 3 \). This expression helps us in finding critical points by setting it to zero and solving for \(x\).

Maximum and Minimum Values

Maximum and minimum values of a function refer to the highest and lowest function outputs on a given interval. After finding the critical points, the next step is to determine where these extremes occur. In our function case, after calculating \( f(x) \) values at these points:

• Calculate \( f(x) \) at critical points obtained from the derivative \((x = 1, -1)\).
• Also include the endpoints \(-\frac{3}{2}\) and \(3\) to ensure we capture all extremes.

From evaluating \( f(x) \) at these values, we found that:

• The minimum value is \( f(1) = -1 \).
• The maximum value is \( f(3) = 19 \).

This tells us where the function reaches its highest and lowest values, confirming that the endpoints can be just as significant as critical points within an interval.

Power Rule in Calculus

The power rule is a fundamental concept for finding derivatives in calculus. It's particularly useful when dealing with polynomial functions. The rule states that to differentiate \( x^n \), the result is \( nx^{n-1} \).
For instance, in the function \( f(x) = x^3 - 3x + 1 \):

• The derivative of \( x^3 \) using the power rule is \( 3x^2 \).
• The \(-3x\) becomes \(-3\) as it is \( x^1 \).
• Constant terms like \(1\) disappear since the derivative of a constant is \(0\).

Understanding this rule is pivotal not only for solving specific problems but also for grasping broader concepts in calculus, as derivatives form the basis for various applications including finding slopes, rates of change, and critical points.