Select The Correct Answer.Over Which Interval Of The Domain Is The Function \[$ H \$\] Decreasing?$\[ H(x) = \begin{cases} 2^x, & X \ \textless \ 1 \\ \sqrt{x+3}, & X \geq 1 \end{cases} \\]A. The Function Is Increasing Only. B.

Introduction

When dealing with piecewise functions, it's essential to understand the behavior of each component function over its respective domain. In this case, we're given a function ${h(x)}$ defined as:

${h(x) = \begin{cases} 2^x, & x \ \textless \ 1 \\ \sqrt{x+3}, & x \geq 1 \end{cases}}$

Our goal is to determine over which interval of the domain the function ${h(x)}$ is decreasing.

Understanding the Component Functions

To analyze the behavior of the function ${h(x)}$, we need to examine each component function separately.

Component Function 1: $2^x$

The first component function is $2^x$, which is an exponential function with base 2. This function is increasing over its entire domain, which is all real numbers. In other words, as $x$ increases, $2^x$ also increases.

Component Function 2: $\sqrt{x+3}$

The second component function is $\sqrt{x+3}$, which is a square root function with a horizontal shift of 3 units to the left. This function is increasing over its domain, which is all real numbers greater than or equal to -3.

Analyzing the Behavior of the Function ${h(x)}$

Now that we've examined each component function, let's analyze the behavior of the function ${h(x)}$ over its domain.

Interval 1: $x < 1$

Over the interval $x < 1$, the function ${h(x)}$ is defined as $2^x$. As we discussed earlier, this function is increasing over its entire domain. Therefore, over the interval $x < 1$, the function ${h(x)}$ is also increasing.

Interval 2: $x \geq 1$

Over the interval $x \geq 1$, the function ${h(x)}$ is defined as $\sqrt{x+3}$. As we discussed earlier, this function is increasing over its domain. However, we need to determine if it's increasing or decreasing over the interval $x \geq 1$.

To do this, we can take the derivative of the function $\sqrt{x+3}$ with respect to $x$. Using the chain rule, we get:

${\frac{d}{dx} \sqrt{x+3} = \frac{1}{2 \sqrt{x+3}}}$

Since the derivative is positive over the interval $x \geq 1$, the function $\sqrt{x+3}$ is increasing over this interval.

Conclusion

Based on our analysis, we can conclude that the function ${h(x)}$ is increasing over the interval $x < 1$ and the interval $x \geq 1$. Therefore, the correct answer is:

A. The function is increasing only.

Discussion

The function ${h(x)}$ is a piecewise function defined over two intervals: $x < 1$ and $x \geq 1$. Over the interval $x < 1$, the function is defined as $2^x$, which is an increasing function. Over the interval $x \geq 1$, the function is defined as $\sqrt{x+3}$, which is also an increasing function.

Therefore, the function ${h(x)}$ is increasing over its entire domain, and the correct answer is A. The function is increasing only.

Final Answer

The final answer is A. The function is increasing only.

Introduction

In our previous article, we analyzed the behavior of the function ${h(x)}$ over its domain and concluded that it is increasing over the interval $x < 1$ and the interval $x \geq 1$. However, we were asked to determine over which interval of the domain the function ${h(x)}$ is decreasing.

In this article, we will provide a Q&A section to help clarify any doubts and provide additional insights into the behavior of the function ${h(x)}$.

Q&A

Q1: What is the domain of the function ${h(x)}$?

A1: The domain of the function ${h(x)}$ is all real numbers, divided into two intervals: $x < 1$ and $x \geq 1$.

Q2: What is the behavior of the function $2^x$ over its domain?

A2: The function $2^x$ is an increasing function over its entire domain, which is all real numbers.

Q3: What is the behavior of the function $\sqrt{x+3}$ over its domain?

A3: The function $\sqrt{x+3}$ is an increasing function over its domain, which is all real numbers greater than or equal to -3.

Q4: Over which interval of the domain is the function ${h(x)}$ decreasing?

A4: The function ${h(x)}$ is not decreasing over any interval of its domain. It is increasing over the interval $x < 1$ and the interval $x \geq 1$.

Q5: Why is the function ${h(x)}$ not decreasing over any interval of its domain?

A5: The function ${h(x)}$ is not decreasing over any interval of its domain because both component functions, $2^x$ and $\sqrt{x+3}$, are increasing functions over their respective domains.

Q6: What is the derivative of the function $\sqrt{x+3}$ with respect to $x$?

A6: The derivative of the function $\sqrt{x+3}$ with respect to $x$ is $\frac{1}{2 \sqrt{x+3}}$.

Q7: Is the derivative of the function $\sqrt{x+3}$ positive or negative over the interval $x \geq 1$?

A7: The derivative of the function $\sqrt{x+3}$ is positive over the interval $x \geq 1$, which means the function $\sqrt{x+3}$ is increasing over this interval.

Conclusion

In this Q&A article, we provided additional insights into the behavior of the function ${h(x)}$ over its domain. We clarified any doubts and provided answers to frequently asked questions. We concluded that the function ${h(x)}$ is not decreasing over any interval of its domain, but rather increasing over the interval $x < 1$ and the interval $x \geq 1$.

Final Answer

The final answer is A. The function is increasing only.