Let's break this down step by step, focusing on each part of the question.

14. $f(g(2))$

Given:

• $f(x) = 2x - 3$
• $g(x) = 4 - x^2$

We first need to evaluate $g(2)$ and then substitute the result into $f(x)$:

1. $g(2) = 4 - (2)^2 = 4 - 4 = 0$
2. $f(g(2)) = f(0) = 2(0) - 3 = -3$

Thus, $f(g(2)) = -3$.

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15. $f(f(1))$

We need to evaluate $f(1)$ first and then substitute that value back into $f(x)$.

1. $f(1) = 2(1) - 3 = 2 - 3 = -1$
2. $f(f(1)) = f(-1) = 2(-1) - 3 = -2 - 3 = -5$

Thus, $f(f(1)) = -5$.

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16. $(f \circ g)(6)$

This is $f(g(6))$. First, we evaluate $g(6)$, then substitute it into $f(x)$.

1. $g(6) = 4 - (6)^2 = 4 - 36 = -32$
2. $f(g(6)) = f(-32) = 2(-32) - 3 = -64 - 3 = -67$

Thus, $(f \circ g)(6) = -67$.

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17. $(f \circ f)(5)$

This is $f(f(5))$. First, evaluate $f(5)$, then substitute it back into $f(x)$.

1. $f(5) = 2(5) - 3 = 10 - 3 = 7$
2. $f(f(5)) = f(7) = 2(7) - 3 = 14 - 3 = 11$

Thus, $(f \circ f)(5) = 11$.

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18. Find the compositions of functions

(a) $f \circ g$

The composition $(f \circ g)(x)$ means $f(g(x))$. We substitute $g(x) = 4 - x^2$ into $f(x) = 2x - 3$:

$(f \circ g)(x) = f(g(x)) = f(4 - x^2) = 2(4 - x^2) - 3 = 8 - 2x^2 - 3 = 5 - 2x^2$

Thus, $(f \circ g)(x) = 5 - 2x^2$.

(b) $g \circ f$

The composition $(g \circ f)(x)$ means $g(f(x))$. We substitute $f(x) = 2x - 3$ into $g(x) = 4 - x^2$:

$(g \circ f)(x) = g(f(x)) = g(2x - 3) = 4 - (2x - 3)^2 = 4 - (4x^2 - 12x + 9) = 4 - 4x^2 + 12x - 9$ $(g \circ f)(x) = -4x^2 + 12x - 5$

Thus, $(g \circ f)(x) = -4x^2 + 12x - 5$.

(c) $f \circ f$

The composition $(f \circ f)(x)$ is $f(f(x))$:

$(f \circ f)(x) = f(f(x)) = f(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9$

Thus, $(f \circ f)(x) = 4x - 9$.

(d) $g \circ g$

The composition $(g \circ g)(x)$ is $g(g(x))$:

$(g \circ g)(x) = g(g(x)) = g(4 - x^2) = 4 - (4 - x^2)^2 = 4 - (16 - 8x^2 + x^4) = 4 - 16 + 8x^2 - x^4$ $(g \circ g)(x) = -12 + 8x^2 - x^4$

Thus, $(g \circ g)(x) = -x^4 + 8x^2 - 12$.

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Domains of the functions

• $f(x) = 2x - 3$: Domain is all real numbers $\mathbb{R}$.
• $g(x) = 4 - x^2$: Domain is all real numbers $\mathbb{R}$.
• $f \circ g$: Domain is all real numbers $\mathbb{R}$ (since both functions are defined for all $x$).
• $g \circ f$: Domain is all real numbers $\mathbb{R}$.
• $f \circ f$: Domain is all real numbers $\mathbb{R}$.
• $g \circ g$: Domain is all real numbers $\mathbb{R}$.

Would you like to go over any part of the solution in more detail?

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Related questions:

1. How do you determine the domain of composite functions?
2. What is the difference between $f \circ g$ and $g \circ f$?
3. Can you find the composition of inverse functions?
4. What is the range of $f(x) = 2x - 3$?
5. How do transformations affect the graph of $f(x)$ and $g(x)$?

Tip: When composing functions, always evaluate the inner function first and then substitute its result into the outer function.