Problem 87 - Calculus Single Variable

Question

In each of Exercises 82-89, use the first derivative to determine the intervals on which the function is increasing and on which the function is decreasing. $$ f(x)=x^{3}-x^{2}+e^{-x} $$

Step-by-Step Solution

Answer

Find where \( f'(x) \) is positive or negative to determine intervals of increase or decrease. The critical points need technology or numerical methods for exact solutions.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first need to find its derivative. The original function is given by: \[ f(x) = x^3 - x^2 + e^{-x} \]Calculate the derivative using the rules for derivatives:\[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(x^2) + \frac{d}{dx}(e^{-x}) \]This gives us:\[ f'(x) = 3x^2 - 2x - e^{-x} \]

Step 2: Solve for Critical Points

Find the critical points by setting the derivative equal to zero:\[ 3x^2 - 2x - e^{-x} = 0 \]This equation often needs numerical methods or graphing to solve, as it involves an exponential function and a polynomial. Assume some critical points exist which can best be found using technology.

Step 3: Determine Signs of the Derivative

Determine the sign of the derivative \( f'(x) \) on the intervals determined by the critical points. Check to the left and right of the critical points by testing values in \( f'(x) = 3x^2 - 2x - e^{-x} \).For instance, pick test points such as \( x = -1 \), \( x = 0 \), \( x = 2 \), etc., and substitute them into \( f'(x) \) to see if it is positive or negative in each interval.

Step 4: Establish Intervals of Increase or Decrease

Based on the sign of \( f'(x) \):- If \( f'(x) > 0 \) for a certain interval, the function is increasing on that interval.- If \( f'(x) < 0 \) for a certain interval, the function is decreasing on that interval.Summarize the intervals based on the tests conducted in Step 3.

Key Concepts

Derivatives

Derivatives are fundamental to calculus and involve measuring the rate at which a function changes at any given point. Imagine them as the mathematical equivalent of answering the question, "How fast?"

When you find the derivative of a function, like the one provided in the original exercise, you essentially determine how the variables within the function are changing relative to each other. Derivatives are represented with notations like \( f'(x) \) or \( \frac{dy}{dx} \), and calculate them using rules such as the power rule, product rule, and chain rule.

For example, in the function \( f(x) = x^3 - x^2 + e^{-x} \), the derivative \( f'(x) = 3x^2 - 2x - e^{-x} \) was calculated by applying these rules. Each part of the function—\( x^3 \), \( x^2 \), and \( e^{-x} \)—was differentiated separately and then combined to create the derivative function. This essence of breaking down a function's components makes derivative calculation manageable and systematic.

Critical Points

Critical points are specific values of \( x \) where the derivative of a function equals zero or is undefined. These points give insights into where the function may change from increasing to decreasing or vice versa.

To find these critical points for a function like \( f(x)=x^{3}-x^{2}+e^{-x} \) with the derivative \( f'(x) = 3x^2 - 2x - e^{-x} \), you would set up the equation \( f'(x) = 0 \) and solve for \( x \).

Finding critical points involves solving sometimes complex equations. As seen in the original solution, solving \( 3x^2 - 2x - e^{-x} = 0 \) can involve a mix of numerical methods or using graphing technology due to the presence of both polynomial and exponential components.

Once you have critical points, they serve as boundaries for determining where the function might experience peaks, valleys, or even plateaus.

Increasing and Decreasing Functions

A function is termed as increasing on an interval if, as \( x \) moves from one point to another in that interval, \( f(x) \) keeps getting larger. Conversely, it is decreasing if \( f(x) \) gets smaller with movement across the interval.

The first derivative, \( f'(x) \), is a handy tool for identifying these intervals. Here’s how:

• If \( f'(x) > 0 \), the function increases on that interval.
• If \( f'(x) < 0 \), the function decreases.

By plugging in values between the critical points into \( f'(x) \), we can ascertain the sign of the derivative over exclusive segments. For example, if testing gives a positive \( f'(x) \) around a certain \( x \), it indicates the function is rising over that segment. If negative, it descends.

This simple sign test is crucial, enabling us to map out on intervals accurately, helping to sketch or predict the function's behavior over a defined domain.