Let f : R - {-1/2} → R and g : R - {-5/2} → R be defined as f(x) = (2x+3)/(2x+1) and g(x) = (|x|+1)/

Q: Wait, why do we need to check $g(x) \neq -\frac{1}{2}$? Isn't it enough to just look at the domain of $g(x)$?

That is a classic trap! If you only look at the domain of $g(x)$, you are only checking the first machine in the assembly line. But if $g(x)$ outputs a value that makes $f(x)$ undefined, the whole process fails. The composite function $f(g(x))$ is only defined if $x$ is in the domain of $g$ AND $g(x)$ is in the domain of $f$. You must respect both filters!

Q: I solved $2|x| + 2 = -2x - 5$ and got $x = -\frac{7}{4}$, but the answer says this isn't a restriction. Did I do something wrong?

You solved the algebra correctly, but you missed the logical constraint! When you assumed $x \ge 0$ to open the absolute value, you found $x = -\frac{7}{4}$. But $-\frac{7}{4}$ is not greater than or equal to zero. This is a contradiction, meaning that specific solution is invalid in that case. Always check your final $x$ against the assumption you made for the absolute value!

Q: Why did the second case ($x < 0$) result in $2 = -5$? Is that a mistake?

Not at all! That is actually a beautiful result. When you get a statement like $2 = -5$, it means the equation has no solution in that domain. It tells us that for all negative values of $x$, the function $g(x)$ will never output $-\frac{1}{2}$. It is a mathematical way of saying 'no entry' for that specific case.

Question

Let $f : \mathbb{R} - \{-\frac{1}{2}\} 	o \mathbb{R}$ and $g : \mathbb{R} - \{-\frac{5}{2}\} 	o \mathbb{R}$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is :

Options:

(A) $\mathbb{R} - \{-\frac{5}{2}\}$

(B) $\mathbb{R}$

(C) $\mathbb{R} - \{-\frac{7}{4}\}$

(D) $\mathbb{R} - \{-\frac{5}{2}, -\frac{7}{4}\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the context of composite functions, we define the system as $f(g(x))$. For this system to function, the input $x$ must be valid for the inner function $g(x)$, and the resulting output $g(x)$ must be a valid input for the outer function $f(x)$.

If either condition is violated, the system fails. We must identify all values of $x$ that satisfy these constraints to determine the valid domain.

### The Two Security Checkpoints

**Checkpoint 1:** The input $x$ must be valid for the inner function $g(x) = \frac{|x|+1}{2x+5}$. Since division by zero is undefined, we require the denominator to be non-zero:
$$2x + 5 
eq 0 \implies x 
eq -\frac{5}{2}$$

**Checkpoint 2:** The output $g(x)$ must be a valid input for $f(x) = \frac{2x+3}{2x+1}$. The function $f(x)$ is undefined when its denominator is zero, specifically when $x = -\frac{1}{2}$.

Therefore, we must ensure that $g(x) 
eq -\frac{1}{2}$ for all $x$ in the domain of $g$. If $g(x)$ ever equals $-\frac{1}{2}$, the composite function $f(g(x))$ will crash.

### The Algebraic Investigation

To find if any values of $x$ cause a system failure at the second checkpoint, we solve the equation:
$$\frac{|x|+1}{2x+5} = -\frac{1}{2}$$

Cross-multiplying to clear the fractions, we obtain:
$$2(|x| + 1) = -(2x + 5)$$
$$2|x| + 2 = -2x - 5$$

Because of the absolute value term $|x|$, we must analyze this equation across two distinct intervals.

### Case Analysis

**Case 1: $x \ge 0$**
In this interval, $|x| = x$. The equation simplifies to:
$$2x + 2 = -2x - 5$$
$$4x = -7 \implies x = -\frac{7}{4}$$

Since our assumption was $x \ge 0$, the value $x = -\frac{7}{4}$ is a contradiction. We must reject this solution.

**Case 2: $x < 0$**
In this interval, $|x| = -x$. The equation becomes:
$$-2x + 2 = -2x - 5$$

Adding $2x$ to both sides results in $2 = -5$. This is a mathematical impossibility, indicating that no value of $x$ in this interval satisfies the equation.

### The Triumph of Logic

Our investigation confirms that $g(x)$ never outputs $-\frac{1}{2}$. Consequently, the second security checkpoint never blocks any input.

The only restriction on the domain remains the one identified at the first checkpoint. The domain of the composite function $f(g(x))$ is all real numbers except for $x = -\frac{5}{2}$.

**Final Domain:** $\mathbb{R} \setminus \{ -\frac{5}{2} \}$

Animated Solution for Mathematics - Functions: Let $f : R - {- \frac{1}{2}} \to R$ and $g : R - {- \frac{5}{2}} \to R$ be defined as $f (x) = \frac{2 x + 3}{2 x + 1}$ and $g (x) = \frac{∣ x ∣ + 1}{2 x + 5}$ . Then the domain of the function $f \circ g$ is :

Select Answer:

Visualized Solution

Concept of Composite Domain

The Two Filters

Filter 1: Domain of $g (x)$

Filter 2: Domain of $f (x)$

Setting up the Constraint

Solving the Equation

Case 1: $x \geq 0$

Case 2: $x < 0$

Synthesizing the Constraints

Final Domain

The Sigma Insight: Composite Functions

Analyzing the Setup

The Two Security Checkpoints

The Algebraic Investigation

Case Analysis

The Triumph of Logic

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Visualized Solution

Concept of Composite Domain

The Two Filters

Filter 1: Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Filter 2: Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Setting up the Constraint

Solving the Equation

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0