1. **Stating the problem:** We have a piecewise function defined as: $$g(x) = \begin{cases} 3^x - \sqrt{x} & \text{for } x \in (0,1) \\ x - 5 + \log_2(x-1) & \text{for } x \in (1,3) \end{cases}$$ The intervals given are (0,1), (1,3), and (3,5), but the function is only defined explicitly on (0,1) and (1,3). 2. **Understanding the function:** - For $x \in (0,1)$, the function is $g(x) = 3^x - \sqrt{x}$. - For $x \in (1,3)$, the function is $g(x) = x - 5 + \log_2(x-1)$. 3. **Important notes:** - The square root $\sqrt{x}$ is defined for $x \geq 0$. - The logarithm $\log_2(x-1)$ is defined for $x-1 > 0 \Rightarrow x > 1$. 4. **Evaluating the function at sample points:** - For $x=0.5$ in $(0,1)$: $$g(0.5) = 3^{0.5} - \sqrt{0.5} = \sqrt{3} - \sqrt{0.5} \approx 1.732 - 0.707 = 1.025$$ - For $x=2$ in $(1,3)$: $$g(2) = 2 - 5 + \log_2(2-1) = 2 - 5 + \log_2(1) = -3 + 0 = -3$$ 5. **Behavior on intervals:** - On $(0,1)$, $3^x$ grows from 1 to 3, and $\sqrt{x}$ grows from 0 to 1, so $g(x)$ is positive and increasing. - On $(1,3)$, $x-5$ is negative for $x<5$, and $\log_2(x-1)$ increases from 0 upwards, so $g(x)$ starts negative and increases. 6. **No explicit definition on $(3,5)$ is given, so $g(x)$ is undefined there.** **Final answer:** The piecewise function is $$g(x) = \begin{cases} 3^x - \sqrt{x} & x \in (0,1) \\ x - 5 + \log_2(x-1) & x \in (1,3) \end{cases}$$ with no definition on $(3,5)$.