Over Which Interval Of The Domain Is The Function $h$ Decreasing?$h(x)=\left\{\begin{array}{ll} 2^x, & X\ \textless \ 1 \\ \sqrt{x+3}, & X \geq 1 \end{array}\right.$A. $(-\infty, \infty$\] B. The Function Is Increasing Only.

Introduction

The function $h$ is defined as a piecewise function, with two different expressions for $x < 1$ and $x \geq 1$. To determine the interval where the function $h$ is decreasing, we need to analyze the behavior of both expressions separately and then combine the results.

Expression for $x < 1$

For $x < 1$, the function $h$ is defined as $h(x) = 2^x$. This is an exponential function with base 2, which is always increasing. In other words, as $x$ increases, $2^x$ also increases. Therefore, the function $h$ is increasing for $x < 1$.

Expression for $x \geq 1$

For $x \geq 1$, the function $h$ is defined as $h(x) = \sqrt{x+3}$. This is a square root function, which is increasing for $x \geq 1$. However, we need to analyze the behavior of this function more carefully to determine if it is increasing or decreasing.

Derivative of the Square Root Function

To determine the behavior of the square root function, we can find its derivative. The derivative of $h(x) = \sqrt{x+3}$ is given by:

$$h'(x) = \frac{1}{2\sqrt{x+3}}$$

Critical Points

To find the critical points of the function, we need to find the values of $x$ where the derivative is equal to zero or undefined. In this case, the derivative is undefined when $x+3 = 0$, which gives $x = -3$. However, this is outside the domain of the function, which is $x \geq 1$. Therefore, there are no critical points in the domain of the function.

Increasing or Decreasing

Since the derivative is always positive for $x \geq 1$, the function $h$ is always increasing for $x \geq 1$.

Combining the Results

We have found that the function $h$ is increasing for $x < 1$ and $x \geq 1$. Therefore, the function $h$ is increasing for all values of $x$ in its domain.

Conclusion

Based on our analysis, we can conclude that the function $h$ is increasing for all values of $x$ in its domain. Therefore, the correct answer is:

• B. The function is increasing only.

Final Answer

The final answer is B.

Q: What is the domain of the function h?

A: The domain of the function h is all real numbers x such that x < 1 or x ≥ 1.

Q: What is the range of the function h?

A: The range of the function h is all real numbers y such that y > 0.

Q: Is the function h continuous?

A: Yes, the function h is continuous for all values of x in its domain.

Q: Is the function h differentiable?

A: Yes, the function h is differentiable for all values of x in its domain.

Q: What is the derivative of the function h?

A: The derivative of the function h is given by:

• For x < 1, h'(x) = 2^x ln(2)
• For x ≥ 1, h'(x) = 1/(2√(x+3))

Q: What is the second derivative of the function h?

A: The second derivative of the function h is given by:

• For x < 1, h''(x) = 2^x (ln(2))^2
• For x ≥ 1, h''(x) = -1/(4(x+3)^(3/2))

Q: Is the function h concave up or concave down?

A: The function h is concave up for x < 1 and concave down for x ≥ 1.

Q: What is the inflection point of the function h?

A: The inflection point of the function h is x = 1.

Q: Is the function h increasing or decreasing?

A: The function h is increasing for all values of x in its domain.

Q: What is the maximum value of the function h?

A: The maximum value of the function h is not defined, as the function h is increasing for all values of x in its domain.

Q: What is the minimum value of the function h?

A: The minimum value of the function h is 0, which occurs when x = -3.

Q: Is the function h periodic?

A: No, the function h is not periodic.

Q: Is the function h even or odd?

A: The function h is neither even nor odd.

Q: Can the function h be inverted?

A: Yes, the function h can be inverted.

Q: What is the inverse function of h?

A: The inverse function of h is given by:

• For y < 1, h^(-1)(y) = ln(y)/ln(2)
• For y ≥ 1, h^(-1)(y) = √(y-3) - 3

Q: Is the inverse function of h continuous?

A: Yes, the inverse function of h is continuous for all values of y in its range.

Q: Is the inverse function of h differentiable?

A: Yes, the inverse function of h is differentiable for all values of y in its range.

Q: What is the derivative of the inverse function of h?

A: The derivative of the inverse function of h is given by:

• For y < 1, (h^(-1))'(y) = 1/(y ln(2))
• For y ≥ 1, (h^(-1))'(y) = 1/(2√(y-3))

Q: What is the second derivative of the inverse function of h?

A: The second derivative of the inverse function of h is given by:

• For y < 1, (h^(-1))''(y) = -1/(y^2 (ln(2))^2)
• For y ≥ 1, (h^(-1))''(y) = -1/(4(y-3)^(3/2))