Composite Functions - Domain

Question 1: Computing a Composite Function (from file image_c61f68.png)

The Question: If $f(x) = \frac{x+5}{x-4}$ and $g(x) = \frac{1}{x+1}$, then find $(f \circ g)(x)$.

Detailed Solution:

1. Set up the composition: We need to find $f(g(x))$. This means we will substitute the entire expression for $g(x)$ into every instance of $x$ in the function $f(x)$.

   $$f(g(x)) = \frac{g(x)+5}{g(x)-4} = \frac{\left(\frac{1}{x+1}\right) + 5}{\left(\frac{1}{x+1}\right) - 4}$$

2. Simplify the complex fraction: The easiest way to simplify this is to multiply the main numerator and the main denominator by the “mini” denominator, which is $(x+1)$.

   • Simplify the numerator: $$(x+1) \cdot \left(\frac{1}{x+1} + 5\right) = (x+1)\frac{1}{x+1} + (x+1)5 = 1 + 5x + 5 = 5x + 6$$
   • Simplify the denominator: $$(x+1) \cdot \left(\frac{1}{x+1} - 4\right) = (x+1)\frac{1}{x+1} - (x+1)4 = 1 - (4x + 4) = 1 - 4x - 4 = -4x - 3$$

3. Combine the simplified parts:

   $$f(g(x)) = \frac{5x+6}{-4x-3}$$

4. Match the format of the options: We can factor out a -1 from the denominator to match the third option.

   $$f(g(x)) = \frac{5x+6}{-(4x+3)} = -\frac{5x+6}{4x+3} = \frac{-5x-6}{4x+3}$$

Final Answer: The composite function is $\frac{-5x-6}{4x+3}$.

Question 2: Finding the Domain of a Composite Function (from file image_c61f68.png)

The Question: If $f(x) = \frac{x+5}{x-4}$ and $g(x) = \frac{1}{x+1}$, then find the domain of the function $(f \circ g)(x)$.

Detailed Solution:

We must apply our two-condition rule.

1. Condition 1: Find the domain of the inner function, $g(x)$.

   • $g(x) = \frac{1}{x+1}$. The denominator cannot be zero.
   • $x+1 \neq 0 \implies x \neq -1$.
   • So, our first restriction is that $x$ cannot be -1.

2. Condition 2: The output of $g(x)$ must be in the domain of the outer function, $f(x)$.

   • First, find the domain of $f(x) = \frac{x+5}{x-4}$. Its denominator cannot be zero, so its domain is all real numbers except for 4.
   • This means the output of $g(x)$ cannot be 4. We must find which $x$-value causes this and exclude it.
   • Set $g(x) = 4$: $$\frac{1}{x+1} = 4$$$$1 = 4(x+1)$$$$1 = 4x + 4$$$$-3 = 4x$$$$x = -\frac{3}{4}$$
   • So, our second restriction is that $x$ cannot be -3/4.

3. Combine the restrictions: The domain is all real numbers except for the values we must exclude from both conditions. The domain is $\mathbb{R} \setminus {-1, -\frac{3}{4}}$.

Final Answer: $\mathbb{R} \setminus {-1, -\frac{3}{4}}$.

Question 3: Correct Statements about the Domain of a Composite Function (from file image_c61f0c.png)

The Question: Which of the following statements is(are) CORRECT?

Detailed Solution:

Let’s evaluate each statement based on the formal definition of the domain of a composite function, $(f \circ g)(x)$.

• “If there exists $x \in \mathbb{R}$ that is not in the domain of a function $f$, then that $x$ will not be in the domain of some composite function $(f \circ g)$.”

  • This statement is not necessarily true and is poorly worded. For example, if $f(x)=\sqrt{x}$ and $g(x)=x^2$, $x=-2$ is not in the domain of $f$, but $x=-2$ is in the domain of $(f \circ g)(x) = \sqrt{x^2} = |x|$. The key restrictions come from the inner function first.

• “The domain of a composite function $(f \circ g)$ is the set of all $x$ such that $x$ is in the domain of a function $f$ and $f(x)$ is in the domain of a function $g$.”

  • FALSE. This reverses the order. It describes the domain of $(g \circ f)(x)$.

• “If there exists $x \in \mathbb{R}$ that is not in the domain of a function $g$, then that $x$ will not be in the domain of some composite function $(f \circ g)$.”

  • TRUE. This is a direct consequence of the definition. To calculate $f(g(x))$, we must first be able to calculate $g(x)$. If $x$ is not a valid input for $g$, the process fails at the first step.

• “The domain of a composite function $(f \circ g)$ is the set of all $x$ such that $x$ is in the domain of a function $g$ and $g(x)$ is in the domain of a function $f$.”

  • TRUE. This is the precise, formal definition that combines our two conditions.

Final Answer: The correct statements are:

• If there exists $x \in \mathbb{R}$ that is not in the domain of a function $g$, then that $x$ will not be in the domain of some composite function $(f \circ g)$.
• The domain of a composite function $(f \circ g)$ is the set of all $x$ such that $x$ is in the domain of a function $g$ and $g(x)$ is in the domain of a function $f$.

Question 4: Finding a Function from a Composite Domain (from file image_c61f0c.png)

The Question: If the domain of a composite function $(f \circ g)$ is $(-\infty, 2)$ and $f(x) = \frac{1}{x^2}$, then which of the following options can be $g(x)$?

Detailed Solution:

Let’s use our two-condition rule in reverse.

1. Analyze the outer function, $f(x)$:

   • $f(x) = \frac{1}{x^2}$.
   • The domain of $f$ is all real numbers except where the denominator is zero. The domain is $x \neq 0$.

2. Apply the two conditions for the domain of $(f \circ g)(x) = f(g(x))$:

   • Condition 1: $x$ must be in the domain of $g(x)$.
   • Condition 2: The output $g(x)$ must be in the domain of $f(x)$, which means $g(x) \neq 0$.

3. We are told the final domain, after applying both conditions, is $(-\infty, 2)$, which means $x < 2$. Let’s test each option for $g(x)$ to see which one results in this domain.

• Option: $g(x) = \sqrt{2-x}$

  • Condition 1 (Domain of g): The expression inside the square root must be non-negative. $2 - x \ge 0 \implies 2 \ge x$, or $x \le 2$.
  • Condition 2 (g(x) must not be 0): We need to find where $g(x)=0$ and exclude that point. $\sqrt{2-x} = 0 \implies 2-x = 0 \implies x = 2$. So we must exclude $x=2$.
  • Combine: The domain of $g(x)$ is $x \le 2$. The restriction from $f(x)$ is $x \neq 2$. The intersection of these two conditions is $x < 2$, which is the interval $(-\infty, 2)$. This is a perfect match.

• Let’s check another option to be sure: $g(x) = \frac{3}{\sqrt{x+2}}$

  • Condition 1 (Domain of g): The expression in the square root must be strictly positive (since it’s also in the denominator). $x+2 > 0 \implies x > -2$.
  • Condition 2 (g(x) must not be 0): The fraction $\frac{3}{\sqrt{x+2}}$ can never be zero. So there is no additional restriction.
  • Combine: The final domain would be $x > -2$, or $(-2, \infty)$, which does not match.

Final Answer: The correct option is $g(x) = \sqrt{2-x}$.