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) $$ f(x)=\sqrt{2 x-1}-3 $$

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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) $$ f(x)=\sqrt{2 x-1}-3 $$

UpStudy AI · Accepted Answer

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.

**Domains:** 1. $$ f(x) = 1 + \sqrt{2x + 1} $$: $$ x \geq -\frac{1}{2} $$ 2. $$ f(x) = \sqrt{x + 3} + 2 $$: $$ x \geq -3 $$ 3. $$ f(x) = 1 + \sqrt{\frac{1}{2}} $$: All real numbers ($$ x \in \mathbb{R} $$) 4. $$ f(x) = 2 + 3\sqrt{-x + 1} $$: $$ x \leq 1 $$ 5. $$ f(x) = 1 - 3\sqrt{x - \frac{1}{2}} $$: $$ x \geq \frac{1}{2} $$ 6. $$ f(x) = \sqrt{2x - 1} - 3 $$: $$ x \geq \frac{1}{2} $$ **Sketches:** Each function is a transformation of the basic square root function $$ y = \sqrt{x} $$, involving shifts, scaling, and reflections based on the coefficients and constants provided. The domains restrict the x-values for which the functions are real and defined.

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) \( f(x)=\sqrt{2 x-1}-3 \)

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