To compute the compositions of functions, $(f \circ g)(x)$ and $(g \circ f)(x)$, follow these steps:

1. Given Functions:

• $f(x) = 3x - 2$
• $g(x) = \frac{x + 1}{2}$

2. Compute $(f \circ g)(x)$:

The composition $(f \circ g)(x)$ means substituting $g(x)$ into $f(x)$.

$(f \circ g)(x) = f(g(x)) = f\left(\frac{x+1}{2}\right)$

Substitute $\frac{x+1}{2}$ into $f(x) = 3x - 2$:

$f\left(\frac{x+1}{2}\right) = 3\left(\frac{x+1}{2}\right) - 2$

Simplify:

$f\left(\frac{x+1}{2}\right) = \frac{3(x+1)}{2} - 2$ $f\left(\frac{x+1}{2}\right) = \frac{3x + 3}{2} - 2$ $f\left(\frac{x+1}{2}\right) = \frac{3x + 3}{2} - \frac{4}{2}$ $f\left(\frac{x+1}{2}\right) = \frac{3x - 1}{2}$

Thus, $(f \circ g)(x) = \frac{3x - 1}{2}$

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3. Compute $(g \circ f)(x)$:

The composition $(g \circ f)(x)$ means substituting $f(x)$ into $g(x)$.

$(g \circ f)(x) = g(f(x)) = g(3x - 2)$

Substitute $3x - 2$ into $g(x) = \frac{x+1}{2}$:

$g(3x - 2) = \frac{(3x - 2) + 1}{2}$

Simplify:

$g(3x - 2) = \frac{3x - 2 + 1}{2}$ $g(3x - 2) = \frac{3x - 1}{2}$

Thus, $(g \circ f)(x) = \frac{3x - 1}{2}$

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Final Results:

1. $(f \circ g)(x) = \frac{3x - 1}{2}$
2. $(g \circ f)(x) = \frac{3x - 1}{2}$

Observations:

The compositions $(f \circ g)(x)$ and $(g \circ f)(x)$ result in the same expression: $\frac{3x - 1}{2}$.

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Would you like further details or a step-by-step explanation of any part?

Related Questions:

1. What is the domain of $f(x)$, $g(x)$, and their compositions?
2. How do you compute $(f \circ f)(x)$ or $(g \circ g)(x)$?
3. Can $f$ and $g$ be inverses of each other? Why or why not?
4. What is the graphical interpretation of $(f \circ g)(x)$ and $(g \circ f)(x)$?
5. How do the compositions change if $g(x) = 2x + 1$ instead?

Tip:

When dealing with compositions, always substitute the inner function carefully and simplify step by step to avoid errors!