Problem 10 - 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. $$ f(x)=x^{3}-3 x+1 ; I=\left[-\frac{3}{2}, 3\right] $$

Step-by-Step Solution

Answer

Max value is 19 at \( x = 3 \), min value is \(-1\) at \( x = 1 \).

Step 1: Find the derivative of the function

To find the critical points, we need to find the derivative of the function. The function given is \( f(x)=x^3 - 3x + 1 \). The derivative, \( f'(x) \), is found using basic differentiation rules:\[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x) + \frac{d}{dx}(1) \]\[ f'(x) = 3x^2 - 3 \].

Step 2: Find critical points

Critical points occur where the derivative is zero or undefined. The derivative we found is \( f'(x) = 3x^2 - 3 \). Set this equal to zero and solve for \( x \):\[ 3x^2 - 3 = 0 \]\[ 3(x^2 - 1) = 0 \]\[ x^2 - 1 = 0 \]\[ (x - 1)(x + 1) = 0 \]So, \( x = 1 \) and \( x = -1 \) are critical points. These points are within the interval \( I = \left[-\frac{3}{2}, 3\right] \).

Step 3: Evaluate the function at the critical points and endpoints

Evaluate the function \( f(x) \) at the critical points \( x = 1 \) and \( x = -1 \), and also at the endpoints of the interval \( I \), which are \( x = -\frac{3}{2} \) and \( x = 3 \).- At \( x = -\frac{3}{2} \): \[ f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^3 - 3\left(-\frac{3}{2}\right) + 1 = -\frac{27}{8} + \frac{9}{2} + 1 = \frac{5}{8} \]- At \( x = -1 \): \[ f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 \]- At \( x = 1 \): \[ f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1 \]- At \( x = 3 \): \[ f(3) = (3)^3 - 3(3) + 1 = 27 - 9 + 1 = 19 \].

Step 4: Determine maximum and minimum values

Based on the evaluations:- \( f\left(-\frac{3}{2}\right) = \frac{5}{8} \)- \( f(-1) = 3 \)- \( f(1) = -1 \)- \( f(3) = 19 \)The maximum value is \( 19 \) at \( x = 3 \), and the minimum value is \( -1 \) at \( x = 1 \).

Key Concepts

Maximum Value

In calculus, finding the maximum value of a function on a specific interval involves identifying the highest point of the function within that range. This can be quite useful in various real-life scenarios, like determining the maximum profit or the highest temperature on a given day.
To find the maximum value, you evaluate the function at its critical points and at the endpoints of the interval. These are the points where the highest value might be found, as the function could reach its peak at these locations.
In the exercise, the function was evaluated at critical points and endpoints within the given interval \(I = \left[-\frac{3}{2}, 3\right]\). After evaluation, the maximum value of the function \(f(x)\) on this interval is \(19\), which occurs at \(x = 3\). This value is significant because it represents the highest output value of the function within the specified interval.

Minimum Value

Finding the minimum value of a function is similar to finding the maximum, but here we look for the lowest point the function reaches within the given interval. This is important for tasks like minimizing costs or finding the lowest time needed for a trip.
Similar to maximum values, we evaluate the function at critical points and endpoints to find the minimum value. The critical points are where the function's behavior might change, leading to potential minimum values.
From the exercise, the minimum value of \(f(x)\) on the interval \(I\) is the lowest output at \(x = 1\), which gives us \(-1\). This point is the least value reached by the function within the interval, highlighting its importance in understanding the behavior of the function.

Derivative

The derivative of a function is a fundamental concept in calculus. It measures how a function changes as its input changes, essentially representing the slope of the function at any point.
To locate critical points where the function might have a maximum or minimum value, we first find its derivative. In this exercise, the function \(f(x) = x^3 - 3x + 1\) has a derivative calculated as \(f'(x) = 3x^2 - 3\).
This derivative equation is crucial because it helps us understand where the function's rate of change is zero or undefined, locations that indicate potential maximum or minimum points. Identifying and solving these equations is an essential step in analyzing the behavior of functions.

Evaluation of Endpoints

Evaluating the endpoints of the given interval is a necessary step in finding the maximum and minimum values of a function within that interval. This is because, at these boundaries, the function might either ascend or descend, potentially presenting a maximum or minimum value.
In the given interval \(I = \left[-\frac{3}{2}, 3\right]\), endpoints are \(x = -\frac{3}{2}\) and \(x = 3\). Calculating the function at these points provides values that need to be compared with those found at any critical points.
- At \(x = -\frac{3}{2}\), \(f(x)\) evaluates to \(\frac{5}{8}\)
- At \(x = 3\), \(f(x)\) evaluates to \(19\)
These evaluations are crucial as they can influence the determination of which points offer the maximum and minimum values in the given interval.