3. A) Sketch the following and find the domains: i) \( f(x)=1+\sqrt{2 x+1} \) ii) \( f(x)=\sqrt{x+3}+2 \) iii) \( f(x)=1+\sqrt{\frac{1}{2}} \) iv) \( f(x)=2+3 \sqrt{-x+1} \) (v) \( f(x)=1-3 \sqrt{x-\frac{1}{2}} \) (vi) \( f(x)=\sqrt{2 x-1}-3 \)

Alright, I need to tackle this problem involving sketching functions and finding their domains. Let's break it down step by step for each function. **i) \( f(x) = 1 + \sqrt{2x + 1} \)** First, I'll consider the expression inside the square root: \( 2x + 1 \). For the square root to be real, the expression inside must be greater than or equal to zero. So, \( 2x + 1 \geq 0 \). Solving for \( x \), I get \( x \geq -\frac{1}{2} \). This means the domain of \( f(x) \) is all real numbers greater than or equal to \( -\frac{1}{2} \). Now, to sketch this function, I'll start by plotting the basic square root function \( y = \sqrt{x} \). Then, I'll shift it up by 1 unit and to the left by \( \frac{1}{2} \) unit. The graph will start at \( x = -\frac{1}{2} \) and extend to the right, increasing as \( x \) increases. **ii) \( f(x) = \sqrt{x + 3} + 2 \)** Similarly, for this function, the expression inside the square root is \( x + 3 \). Setting \( x + 3 \geq 0 \) gives \( x \geq -3 \). So, the domain is all real numbers greater than or equal to \( -3 \). To sketch, I'll plot \( y = \sqrt{x} \), shift it up by 2 units, and to the left by 3 units. The graph will start at \( x = -3 \) and extend to the right, increasing as \( x \) increases. **iii) \( f(x) = 1 + \sqrt{\frac{1}{2}} \)** Wait a minute, this function seems a bit different. The expression inside the square root is a constant, \( \frac{1}{2} \). Since \( \frac{1}{2} \) is positive, the square root is real. Therefore, the domain is all real numbers, \( x \in \mathbb{R} \). However, the function simplifies to \( f(x) = 1 + \sqrt{\frac{1}{2}} \), which is a constant function. Its graph is a horizontal line at \( y = 1 + \sqrt{\frac{1}{2}} \). There's no variation with \( x \), so the domain is all real numbers. **iv) \( f(x) = 2 + 3\sqrt{-x + 1} \)** Here, the expression inside the square root is \( -x + 1 \). For the square root to be real, \( -x + 1 \geq 0 \), which simplifies to \( x \leq 1 \). So, the domain is all real numbers less than or equal to 1. To sketch, I'll plot \( y = \sqrt{x} \), scale it vertically by 3, shift it up by 2 units, and reflect it over the y-axis (since the expression is \( -x \)). The graph will start at \( x = 1 \) and extend to the left, decreasing as \( x \) decreases. **v) \( f(x) = 1 - 3\sqrt{x - \frac{1}{2}} \)** The expression inside the square root is \( x - \frac{1}{2} \). For the square root to be real, \( x - \frac{1}{2} \geq 0 \), so \( x \geq \frac{1}{2} \). The domain is all real numbers greater than or equal to \( \frac{1}{2} \). To sketch, I'll plot \( y = \sqrt{x} \), scale it vertically by 3, shift it down by 1 unit, and shift it to the right by \( \frac{1}{2} \) unit. The graph will start at \( x = \frac{1}{2} \) and extend to the right, increasing as \( x \) increases. **vi) \( f(x) = \sqrt{2x - 1} - 3 \)** The expression inside the square root is \( 2x - 1 \). Setting \( 2x - 1 \geq 0 \) gives \( x \geq \frac{1}{2} \). So, the domain is all real numbers greater than or equal to \( \frac{1}{2} \). To sketch, I'll plot \( y = \sqrt{x} \), scale it vertically by 2, shift it down by 3 units, and shift it to the right by \( \frac{1}{2} \) unit. The graph will start at \( x = \frac{1}{2} \) and extend to the right, increasing as \( x \) increases. **Summary of Domains:** - i) \( x \geq -\frac{1}{2} \) - ii) \( x \geq -3 \) - iii) All real numbers (\( x \in \mathbb{R} \)) - iv) \( x \leq 1 \) - v) \( x \geq \frac{1}{2} \) - vi) \( x \geq \frac{1}{2} \) **Sketching the Functions:** Each function involves a square root, which introduces a restriction on the domain. The constants and coefficients affect the vertical scaling and shifting of the basic square root graph. Reflecting over the y-axis in some cases changes the direction of the graph's increase or decrease.