Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, identify the critical points and find the maximum value and minimum value on the given interval. $$ f(x)=x^{3}-3 x+1 ; I=\left(-\frac{3}{2}, 3\right) \text { Hint: Sketch the graph. } $$

Step-by-Step Solution

Answer

The maximum value is 19 at \(x = 3\), and the minimum value is -1 at \(x = 1\).

Step 1: Find the derivative

To find the critical points of the function, we first need to find its derivative. The function given is \(f(x) = x^{3} - 3x + 1\). Differentiating with respect to \(x\), we get: \(f'(x) = 3x^{2} - 3\).

Step 2: Set derivative to zero and solve

The critical points occur where the derivative is zero or undefined. In this case, set \(f'(x) = 0\) to find these points: \[3x^{2} - 3 = 0\] Solve for \(x\): 1. \(3(x^{2} - 1) = 0\) 2. \((x - 1)(x + 1) = 0\) Thus, \(x = 1\) and \(x = -1\).

Step 3: Evaluate function at critical points and endpoints

We evaluate \(f(x)\) at the critical points and endpoints of the interval \(I = \left(-\frac{3}{2}, 3\right)\): 1. \(f(-1) = (-1)^3 - 3(-1) + 1 = 3\)2. \(f(1) = 1^3 - 3(1) + 1 = -1\)Evaluate \(f(x)\) at endpoints:3. \(f(-\frac{3}{2}) = \left(-\frac{3}{2}\right)^3 - 3\left(-\frac{3}{2}\right) + 1 = \frac{-15}{8} + \frac{9}{2} + 1 = \frac{25}{8}\)4. \(f(3) = 3^3 - 3(3) + 1 = 19\)Now compare these values to find the maximum and minimum.

Step 4: Identify maximum and minimum values

Among the values calculated:- \(f(-1) = 3\)- \(f(1) = -1\)- \(f(-\frac{3}{2}) = \frac{25}{8}\)- \(f(3) = 19\)The maximum value is \(19\) (at \(x = 3\)), and the minimum value is \(-1\) (at \(x = 1\)).

Key Concepts

Derivative

The derivative of a function gives us important information on how the function behaves. It's essentially a tool that tells us the rate at which a function is changing at any point. This concept is crucial for identifying critical points, which are points where the function may reach a maximum or minimum value within a given interval.
For the function \( f(x) = x^3 - 3x + 1 \), the derivative is calculated as \( f'(x) = 3x^2 - 3 \). Here’s how we determine it:

• Differentiate each term separately: \( x^3 \) becomes \( 3x^2 \), and \( -3x \) becomes \( -3 \).
• The constant term \( +1 \) vanishes, as the derivative of a constant is zero.

This gives us a new function, \( f'(x) \), which we use to find critical points by solving \( f'(x) = 0 \). The solutions to this equation indicate potential maximum or minimum points of \( f(x) \).

Maximum Value

Finding the maximum value of a function within a certain interval involves evaluating the function at critical points and endpoints of the interval. Once potential points are identified through the derivative, each is tested to determine the output value of the original function.
In this case, we need to evaluate \( f(x) \) at the points found and at the interval boundaries:

• At \( x = -1 \), \( f(-1) = 3 \).
• At \( x = 3 \), \( f(3) = 19 \), which is higher than at any other tested point or boundary.

Thus, the highest output value in the interval \((-\frac{3}{2}, 3)\) is 19, making it the maximum value of the function in this case.

Minimum Value

The minimum value of a function is found similarly to the maximum value. After discovering critical points, we evaluate these points alongside the interval’s endpoints to find where the function takes on its smallest value.
For \( f(x) = x^3 - 3x + 1 \), evaluate:

• At \( x = 1 \), \( f(1) = -1 \), which is lower than at any other tested point or boundary.

This evaluation shows that \( f(x) \) reaches its minimum value of \(-1\) at \( x = 1 \), making it the smallest output within the interval \((-\frac{3}{2}, 3)\).

Interval Analysis

Interval analysis involves examining a function over a specific range of values. By assessing the behavior of the function at critical points and endpoints within this interval, we gain a complete understanding of its maximum and minimum values.
For \( f(x) \), we analyze over \( \left(-\frac{3}{2}, 3\right) \):

• Endpoints are \( x = -\frac{3}{2} \) and \( x = 3 \).
• Critical points derived from the derivative are \( x = -1 \) and \( x = 1 \).

By calculating \( f(x) \) at these critical values and points, we identify where the function achieves its highest and lowest outputs within the specified interval. This comprehensive analysis confirms the maximum value is 19 at \( x = 3 \) and the minimum value is \(-1\) at \( x = 1 \), ensuring we understand how \( f(x) \) behaves from start to finish.