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Skills Graph each piecewise-defined function in Exercises \(9-20 .\) Is \(f\) continuous on its domain? Do not use a calculator. $$f(x)=\left\\{\begin{array}{ll} x^{3}+5 & \text { if } x \leq 0 \\ -x^{2} & \text { if } x>0 \end{array}\right.$$

Short Answer

Expert verified
The function is not continuous at \(x = 0\).

Step by step solution

01

Analyze each piece of the function

The function is piecewise, which means it has two different formulas depending on the value of \(x\). Analyze each piece separately: 1. For \(x \leq 0\), the function is \(f(x) = x^3 + 5\). 2. For \(x > 0\), the function is \(f(x) = -x^2\).
02

Graph the first piece

Plotting the first piece \(f(x) = x^3 + 5\) for \(x \leq 0\):- This is a cubic function shifted upwards by 5 units. - On the graph, sketch a cubic curve starting from the left and ending at the point \((0, 5)\). This curve will be included at \(x = 0\) since the inequality \(x \leq 0\) includes 0.
03

Graph the second piece

Plotting the second piece \(f(x) = -x^2\) for \(x > 0\):- This is a downward-opening parabola starting from the point just to the right of \((0, 0)\), as it is not defined at zero but just the points greater than 0.- Sketch this parabola from just to the right of \(x = 0\) downwards as \(x\) increases. Make sure to leave an open circle at \((0, 0)\) as this value is not included in the function.
04

Determine continuity

To check whether the function is continuous on its domain, examine the limits at the point where the function switches, at \(x = 0\):1. The limit of the first piece as \(x\) approaches 0 from the left is \(f(x) = 0^3 + 5 = 5\).2. The limit of the second piece as \(x\) approaches 0 from the right is \(f(x) = -(0)^2 = 0\).Since the left-hand limit (5) does not equal the right-hand limit (0), the function \(f(x)\) is not continuous at \(x = 0\).
05

Finalize the Graph and Conclusions

Combine the two parts of the graph making clear distinctions: - At \(x = 0\), place a solid dot at \((0, 5)\) for the first piece. - Begin the second piece just right of \(x = 0\) with an open circle at \((0, 0)\) moving downward in a parabola.Since there is a jump from the first to the second piece at \(x = 0\), the function is not continuous at this point within its domain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity
In mathematics, the concept of continuity is very important. It tells us how smooth a function is as we move along its graph. A function is continuous if there are no breaks, jumps, or holes in its graph. For a function to be continuous, the left-hand limit, right-hand limit, and the value of the function at that point must all be equal.

In our piecewise function, at the point where the pieces meet (\( x = 0 \)), we need to check for continuity. As calculated, the left-hand limit is 5 and the right-hand limit is 0. Since these values are not the same, there is a discontinuity at \( x = 0 \). This indicates a break or jump in the graph, which means our function is not continuous at this point.
Cubic Function
Cubic functions are an important type of polynomial function, characterized by the highest exponent of 3. Consider the equation \( f(x) = x^3 + 5 \). This is a cubic function shifted up by 5 units because of the '+5'. The graph of a cubic function generally has a curve that can bend and twist in a smoother manner compared to straight lines or simple parabolas.

For the piece of our function defined by \( x \leq 0 \), its graph will start from negative infinity on the left and move eastward, stopping at the point \((0, 5)\). This section of the graph helps show how a cubic function behaves when restricted to this part of the domain.
Parabola
A parabola is the graph of a quadratic function, typically shaped like a U or an upside-down U. In this exercise, the second piece of our piecewise function is \( f(x) = -x^2 \). Here, it represents an upside-down parabola since the \(-x^2\) term causes the graph to open downward.

For \( x > 0 \), the parabola starts just to the right of the y-axis at a point that is not included, depicted with an open circle at \((0,0)\). As \( x \) increases, the graph continues to sweep downwards. This demonstrates how changing the sign in front of \( x^2 \) inverts the parabola direction.
Domain Analysis
Domain analysis involves determining all possible inputs for which a function is defined. In piecewise functions, it often involves checking each 'piece' of the function to ensure there are no restrictions such as divisions by zero or square roots of negative numbers.

For our function, each piece is defined clearly across its stated interval. \( x^3 + 5 \) is valid for all \( x \leq 0 \), while \( -x^2 \) is valid for \( x > 0 \). These intervals make up the entirety of the real number line, excluding the anomaly at \( x = 0 \). Analyzing these domains helps identify where the function might be discontinuous or have other interesting behavior.

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Most popular questions from this chapter

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) ) for each function \(f\). Simplify completely. \(f(x)=\sqrt{x}\) (Hint: Rationalize the numerator.)

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