Select The Correct Answer.Which Function Is Continuous Across Its Domain?A. ${ F(x) = \begin{cases} x-2, & -4 \leq X \ \textless \ -2 \ 0.5x^2, & -2 \leq X \ \textless \ 4 \ 25-3x, & 4 \leq X \leq 8 \end{cases} }$B. $[ F(x) =

Introduction

In mathematics, a function is considered continuous if it can be drawn without lifting the pencil from the paper. This means that the function has no gaps or jumps, and its graph is a single, unbroken curve. However, when dealing with piecewise functions, things can get a bit more complicated. In this article, we will explore the concept of continuous functions and determine which of the given piecewise functions is continuous across its domain.

What is a Continuous Function?

A continuous function is a function that can be drawn without lifting the pencil from the paper. This means that the function has no gaps or jumps, and its graph is a single, unbroken curve. In other words, a continuous function is one where the value of the function at a point is equal to the limit of the function as it approaches that point.

Piecewise Functions

A piecewise function is a function that is defined by multiple sub-functions, each of which is defined on a specific interval. In other words, a piecewise function is a function that is composed of multiple functions, each of which is defined on a specific part of the domain.

Example Piecewise Functions

Let's consider two example piecewise functions:

A. ${ f(x) = \begin{cases} x-2, & -4 \leq x \ \textless \ -2 \ 0.5x^2, & -2 \leq x \ \textless \ 4 \ 25-3x, & 4 \leq x \leq 8 \end{cases} }$

B. ${ f(x) = \begin{cases} x^2, & -2 \leq x \ \textless \ 2 \ 3x-5, & 2 \leq x \leq 5 \end{cases} }$

Which Function is Continuous Across its Domain?

To determine which of the two piecewise functions is continuous across its domain, we need to examine each function separately.

Function A

Let's examine function A:

{ f(x) = \begin{cases} x-2, & -4 \leq x \ \textless \ -2 \\ 0.5x^2, & -2 \leq x \ \textless \ 4 \\ 25-3x, & 4 \leq x \leq 8 \end{cases} \}

To determine if this function is continuous across its domain, we need to examine the behavior of the function at each of the transition points.

• At $x=-2$, we have $\lim_{x\to-2^-}f(x)=\lim_{x\to-2^-}(x-2)=-4$ and $\lim_{x\to-2^+}f(x)=\lim_{x\to-2^+}0.5x^2=2$. Since these limits are not equal, the function is not continuous at $x=-2$.
• At $x=4$, we have $\lim_{x\to4^-}f(x)=\lim_{x\to4^-}0.5x^2=8$ and $\lim_{x\to4^+}f(x)=\lim_{x\to4^+}(25-3x)=17$. Since these limits are not equal, the function is not continuous at $x=4$.

Since the function is not continuous at both $x=-2$ and $x=4$, it is not continuous across its domain.

Function B

Let's examine function B:

{ f(x) = \begin{cases} x^2, & -2 \leq x \ \textless \ 2 \\ 3x-5, & 2 \leq x \leq 5 \end{cases} \}

To determine if this function is continuous across its domain, we need to examine the behavior of the function at each of the transition points.

• At $x=2$, we have $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$ and $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

However, we can see that the function is continuous at $x=2$ if we consider the left-hand and right-hand limits separately. The left-hand limit is $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4$, and the right-hand limit is $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-5)=1$. Since these limits are not equal, the function is not continuous at $x=2$.

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Q&A: Continuous Functions

Q: What is a continuous function?

A: A continuous function is a function that can be drawn without lifting the pencil from the paper. This means that the function has no gaps or jumps, and its graph is a single, unbroken curve.

Q: What is a piecewise function?

A: A piecewise function is a function that is defined by multiple sub-functions, each of which is defined on a specific interval. In other words, a piecewise function is a function that is composed of multiple functions, each of which is defined on a specific part of the domain.

Q: How do I determine if a piecewise function is continuous across its domain?

A: To determine if a piecewise function is continuous across its domain, you need to examine the behavior of the function at each of the transition points. This involves taking the left-hand and right-hand limits of the function at each transition point and checking if they are equal.

Q: What is the difference between a left-hand limit and a right-hand limit?

A: A left-hand limit is the limit of a function as it approaches a point from the left, while a right-hand limit is the limit of a function as it approaches a point from the right.

Q: How do I calculate a left-hand limit and a right-hand limit?

A: To calculate a left-hand limit, you need to approach the point from the left by taking values of the function that are less than the point. To calculate a right-hand limit, you need to approach the point from the right by taking values of the function that are greater than the point.

Q: What happens if the left-hand and right-hand limits are not equal?

A: If the left-hand and right-hand limits are not equal, then the function is not continuous at that point.

Q: Can a piecewise function be continuous across its domain?

A: Yes, a piecewise function can be continuous across its domain if the left-hand and right-hand limits at each transition point are equal.

Q: How do I determine if a piecewise function is continuous across its domain?

A: To determine if a piecewise function is continuous across its domain, you need to examine the behavior of the function at each of the transition points. This involves taking the left-hand and right-hand limits of the function at each transition point and checking if they are equal.

Q: What is the importance of continuity in mathematics?

A: Continuity is an important concept in mathematics because it allows us to study functions in a more rigorous and precise way. It also has many practical applications in fields such as physics, engineering, and economics.

Q: Can you give an example of a continuous function?

A: Yes, the function $f(x) = x^2$ is a continuous function. It is defined for all real numbers and has no gaps or jumps in its graph.

Q: Can you give an example of a discontinuous function?

A: Yes, the function $f(x) = \begin{cases} x-2, & -4 \leq x \ \textless \ -2 \\ 0.5x^2, & -2 \leq x \ \textless \ 4 \\ 25-3x, & 4 \leq x \leq 8 \end{cases}$ is a discontinuous function. It has gaps and jumps in its graph at the transition points.

Conclusion

In conclusion, continuity is an important concept in mathematics that allows us to study functions in a more rigorous and precise way. A piecewise function can be continuous across its domain if the left-hand and right-hand limits at each transition point are equal. We can determine if a piecewise function is continuous across its domain by examining the behavior of the function at each of the transition points.