select the correct answer. over which interval of the domain is function h decreasing? h(x) = {2^x, x < 1…

select the correct answer. over which interval of the domain is function h decreasing? h(x) = {2^x, x < 1; √(x + 3), x ≥ 1. o a. the function is increasing only. o b. (1, ∞) o c. (-∞, ∞) o d. (-∞, 1)

select the correct answer. over which interval of the domain is function h decreasing? h(x) = {2^x, x < 1; √(x + 3), x ≥ 1. o a. the function is increasing only. o b. (1, ∞) o c. (-∞, ∞) o d. (-∞, 1)

Answer

Explanation:

Step1: Analyze $y = 2^{x}$ for $x<1$

The exponential - function $y = a^{x}$ with $a>1$ (here $a = 2$) is an increasing function. So, $y = 2^{x}$ is increasing for $x<1$.

Step2: Analyze $y=\sqrt{x + 3}$ for $x\geq1$

The derivative of $y=\sqrt{x + 3}=(x + 3)^{\frac{1}{2}}$ using the power - rule $\frac{d}{dx}(u^{n})=nu^{n - 1}\frac{du}{dx}$ where $u=x + 3$ and $n=\frac{1}{2}$. The derivative $y^\prime=\frac{1}{2}(x + 3)^{-\frac{1}{2}}\times1=\frac{1}{2\sqrt{x + 3}}$. For $x\geq1$, $y^\prime=\frac{1}{2\sqrt{x + 3}}>0$. So, $y=\sqrt{x + 3}$ is increasing for $x\geq1$.

Answer:

A. The function is increasing only.