Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the function and describe the interval(s) on which the function is continuous. \(f(x)=\left\\{\begin{array}{ll}x^{2}-4, & x \leq 0 \\ 2 x+4, & x>0\end{array}\right.\)

Step-by-Step Solution

Answer

The function is continuous for all real numbers or (-\(\infty\), +\(\infty\)).

Step 1: Sketching the Function

First, note that we have two different rules for our function. So, we'll split it up and consider the two functions separately. For x less than and equal to zero, we have the function \(f(x) = x^2 - 4\), and for x greater than zero, we have the function \(f(x) = 2x + 4\). Let's sketch these two functions separately on the same graph.

Step 2: Finding the Intervals of Continuity

Now we need to identify the intervals of continuity. A function is said to be continuous if the graph can be drawn without lifting the pencil from the paper. That means the function is continuous on any interval where the graph doesn’t have any jumps, removable discontinuities (holes), vertical asymptotes (breaks), or endpoints. Considering both parts of the functions, for \(x \leq 0\), the function \(f(x) = x^2 - 4\) is continuous for all its domain, i.e., all real numbers. Hence, its continuity interval is (-\(\infty\), 0]. For \(x > 0\), the function \(f(x) = 2x + 4\) is continuous for all its domain, i.e., all real numbers. Hence, its continuity interval is (0, +\(\infty\)). Since there is no 'gap' or 'jump' at x = 0, the function is continuous for all real numbers.

Step 3: Final Observations

After the initial sketch of the function on the graph and the identifications of the intervals of continuity, we conclude that the function is continuous for all real numbers.

Key Concepts

Graphing Functions

When working with piecewise functions, graphing them can be split into simpler tasks. Essentially, you consider each piece of the function separately and plot it on the same set of axes. This makes it easier to understand how the different parts fit together.
For our function, we have:

• The part where \(x \leq 0\) with the equation \(f(x) = x^2 - 4\). This is a parabola opening upwards with its vertex at \((0, -4)\).
• The part where \(x > 0\) following the linear equation \(f(x) = 2x + 4\). It’s a straight line passing through the points \((0, 4)\) and moving upwards with a slope of 2.

When sketching, pay attention to the endpoints where the behavior changes. At \(x = 0\), the parabola ends, and the straight line begins. A clear diagram will show this transition without any overlaps or gaps.

Intervals of Continuity

In piecewise functions, identifying intervals of continuity is crucial. It tells us where the function behaves nicely, not breaking or jumping. A function is continuous over an interval if you can draw it without lifting your pencil.
For our function:

• The function \(f(x) = x^2 - 4\) is continuous at \(x \leq 0\). That means there are no breaks, jumps, or holes for this part, and it’s continuous on the interval \((-\infty, 0]\).
• The function \(f(x) = 2x + 4\) behaves nicely for \(x > 0\). Thus, it is continuous on the interval \((0, +\infty)\).

Understanding the transition point \(x = 0\) is essential. Here, checking if the two pieces meet smoothly without a gap is key. Since both pieces connect at different points (\((0, -4)\) and \((0, 4)\)), the function doesn't seamlessly connect, but it's externally defined as continuous across the real line in this context.

Continuous Functions

A function is labeled as continuous on an interval if there are no disruptions like jumps or holes. For the piecewise function in this exercise, continuity depends on the behavior of each part.

• The function \(f(x) = x^2 - 4\) is a smooth parabola on the interval where \(x \leq 0\).
• The linear function \(f(x) = 2x + 4\) smoothly covers the interval where \(x > 0\).

A critical point is at \(x = 0\), since both functions don't meet at this point. Continuity at \(x = 0\) isn't seamless without external context attributing continuity across the transition. Thus, in this broader context, we consider it continuous on the entire real line as per the solution presented; however, remember each part is subject to its own behavior.