Problem 12 - The Calculus with Analytic Geometry

Question

\(f(x)= \begin{cases}-x^{3} & \text { if } x<0 \\ x^{3} & \text { if } x \geq 0\end{cases}\)

Step-by-Step Solution

Answer

The function is continuous, defined as \(f(x) = -x^3\) for \(x < 0\) and \(f(x) = x^3\) for \(x \geq 0\).

Step 1: Identify the Domain Intervals

Notice that the function is defined piecewise with two different expressions depending on the value of x. Specifically, the domain is split into two intervals: \ 1. If \(x < 0\): use the expression \(-x^3\) \ 2. If \(x \geq 0\): use the expression \(x^3\)

Step 2: Evaluate the Function for Sample Values

Evaluate \(f(x)\) at some specific values within each interval to understand the behavior of the function. \ For \(x < 0\), let's choose \(x = -1\): \ \(f(-1) = -(-1)^3 = 1\) \ For \(x \geq 0\), let's choose \(x = 1\): \ \(f(1) = (1)^3 = 1\)

Step 3: Verify Continuity at \(x = 0\)

To ensure the function is continuous, check if the left-hand limit (as \(x\) approaches 0 from the left) equals the right-hand limit (as \(x\) approaches 0 from the right) and the function value at \(x = 0\). \ Left-hand limit (as \(x \to 0^-\)): \ \(\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} -x^3 = 0\) \ Right-hand limit (as \(x \to 0^+\)): \ \(\lim_{{x \to 0^+}} f(x) = \lim_{{x \to 0^+}} x^3 = 0\) \ Function value: \(f(0) = 0^3 = 0\). \ Since all three are equal, the function is continuous at \(x = 0\).

Step 4: Summarize the Function Characteristics

To summarize, the function \(f(x)\) is defined piecewise: \ 1. For \(x < 0\), \(f(x) = -x^3\). \ 2. For \(x \geq 0\), \(f(x) = x^3\). \ The function is continuous at \(x = 0\).

Key Concepts

Domain Intervals

Understanding the domain intervals of a piecewise function is the first step in working with it. The domain of a function consists of all possible input values for which the function is defined. In a piecewise function, the domain is typically divided into intervals where different formulas apply.

For the function given in the exercise, \[f(x)= \begin{cases}-x^{3} & x<0 \ x^{3} & x \geq 0\end{cases}\] the domain is split into two intervals:

• For values where x < 0, the expression \[-x^3\] is used.
• For values where x \geq 0, the expression \[x^3\] is used.

Identifying these intervals helps in understanding how the function behaves in different ranges of x.

Evaluating Piecewise Functions

Evaluating a piecewise function involves plugging in values into the specific formula defined for the interval that the input value (x) falls under. Let's break down the process:

• First, determine which interval the input value belongs to.
• Then, use the corresponding formula to calculate the function's value.

For instance, based on the exercise:

For x < 0, if we choose x = -1: \[f(-1) = -(-1)^3 = 1\]
For x \geq 0, if we choose x = 1: \[f(1) = (1)^3 = 1\]
Evaluating different points in each interval gives a better grasp of the function's overall behavior.

Continuity of Functions

A function is continuous if there are no breaks, jumps, or holes in its graph. For a piecewise function, you must check the boundaries where the different formulas meet. Specifically, we check:

• The left-hand limit as x approaches the boundary from the left.
• The right-hand limit as x approaches the boundary from the right.
• The actual value of the function at the boundary.

For the given function, we need to check continuity at x = 0:

• Left-hand limit as x approaches 0 from the left: \[\text{lim}_{{x \rightarrow 0^-}} f(x) = \text{lim}_{{x \rightarrow 0^-}} -x^3 = 0\]
• Right-hand limit as x approaches 0 from the right: \[\text{lim}_{{x \rightarrow 0^+}} f(x) = \text{lim}_{{x \rightarrow 0^+}} x^3 = 0\]
• Function value at x = 0: \[f(0) = 0^3 = 0\]

Since all these are equal, the function is continuous at x = 0. This ensures there is no interruption in the graph of the function at that point.