Question

$f \circ g(x)$ is equal to:

• A. $1$
• B. $g \circ f(x)$
• C. $\frac{15x + 9}{16x - 5}$
• D. $\frac{1}{x}$
• A. $-3$
• B. $\frac{1}{4}$
• C. $-4$
• D. None of these
• A. $x$
• B. $x^2$
• C. $\frac{5x + 3}{4x - 1}$
• D. $\frac{(x + 3)(5x + 3)}{(4x - 5)(4x - 1)}$
• A. $x$
• B. $x^2$
• C. $2x + 3$
• D. $\frac{x + 3}{4x - 5}$

Answer 1

The Correct Option is B

We are given: \[ f(x) = 2x + 3, \quad g(x) = \frac{x - 3}{2} \] We need to find $f(g(x))$, which by definition means: \[ f(g(x)) = 2 \cdot g(x) + 3 \] Substitute $g(x) = \frac{x - 3}{2}$: \[ f(g(x)) = 2 \cdot \frac{x - 3}{2} + 3 \] The $2$ in numerator and denominator cancel: \[ f(g(x)) = (x - 3) + 3 \] \[ f(g(x)) = x \] So $f(g(x))$ is just $x$. Now, let us compute $g(f(x))$: \[ g(f(x)) = \frac{f(x) - 3}{2} = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x \] We see that $f \circ g(x) = g \circ f(x) = x$. Therefore, $f \circ g(x)$ is equal to $g \circ f(x)$.

Answer 2

We are given: \[ f(x) = 2x + 3, \quad g(x) = \frac{x - 3}{2} \] The condition is: \[ f(x) = g(x - 3) \] Step 1 — Left-hand side: \[ f(x) = 2x + 3 \] Step 2 — Right-hand side: Replace $x$ by $(x - 3)$ in $g(x)$: \[ g(x - 3) = \frac{(x - 3) - 3}{2} = \frac{x - 6}{2} \] Step 3 — Equation: \[ 2x + 3 = \frac{x - 6}{2} \] Multiply through by $2$: \[ 4x + 6 = x - 6 \] \[ 4x - x = -6 - 6 \] \[ 3x = -12 \] \[ x = -4 \] Wait, but -4 is an option (C), not (B). Let’s check calculation — oh, correct, the computed answer is $-4$. So option (C) is correct, not (B).

Answer 3

We already know from Q87 that $f$ and $g$ are inverses, meaning: \[ f \circ g = g \circ f = x \] Let’s compute $(g \circ f)$ once: \[ g(f(x)) = x \] Similarly $(f \circ g)(x) = x$.
Now $(g \circ f \circ g \circ f \circ g \circ f \circ g \circ f)(x)$ — notice that $(g \circ f)$ appears four times. Since each $(g \circ f)$ equals $x$, the whole composition equals $x$.
Similarly, $(f \circ g \circ f \circ g)(x)$ has $(f \circ g)$ twice, which also equals $x$.
Thus: \[ \text{First part} = x, \quad \text{Second part} = x \] Multiplying: \[ x \cdot x = x^2 \] Hence the answer is $x^2$.

Answer 4

We simplify from the inside: - $(f \circ g)(x) = x$ (since $f$ and $g$ are inverses)
- $(g \circ f)(x) = x$ (also true)
So inside the big composition: \[ f \circ (f \circ g) \circ (g \circ f)(x) = f \circ (f \circ g)(x) \] But $(f \circ g)(x) = x$, so this is just: \[ f(x) \] And $f(x) = 2x + 3$. Thus the answer is $2x + 3$.