Let's go through each question one by one from the image and solve them accordingly:

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1. Given functions $f(x) = 2x + 2$ and $g(x) = x^2 - 3$:

a) Find $(f \circ g)(x)$ (composition of $f$ and $g$):

• $(f \circ g)(x)$ means $f(g(x))$.

  $f(g(x)) = f(x^2 - 3) = 2(x^2 - 3) + 2 = 2x^2 - 6 + 2 = 2x^2 - 4$

  So, $(f \circ g)(x) = 2x^2 - 4$.

b) Find $(g \circ f)(x)$ (composition of $g$ and $f$):

• $(g \circ f)(x)$ means $g(f(x))$.

  $g(f(x)) = g(2x + 2) = (2x + 2)^2 - 3 = (4x^2 + 8x + 4) - 3 = 4x^2 + 8x + 1$

  So, $(g \circ f)(x) = 4x^2 + 8x + 1$.

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2. Explain why $(f \circ g)(x)$ is not always the same as $(g \circ f)(x)$, and give an example:

• The composition of functions depends on the order of operations. When composing $f$ and $g$, the operations performed on $x$ are different. For example, in $(f \circ g)(x)$, the function $g(x)$ is applied first, followed by $f$, while in $(g \circ f)(x)$, $f(x)$ is applied first.

Example: From question 1, we have:

• $(f \circ g)(x) = 2x^2 - 4$
• $(g \circ f)(x) = 4x^2 + 8x + 1$

Clearly, $2x^2 - 4 \neq 4x^2 + 8x + 1$, so the two compositions are not the same.

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3. Given $f(x) = 2x + 3$ and $g(x) = x - 1$, find $(f \circ g)^{-1}(x)$ (the inverse of the composition of $f$ and $g$):

1. First, find $(f \circ g)(x)$:

   $(f \circ g)(x) = f(g(x)) = f(x - 1) = 2(x - 1) + 3 = 2x - 2 + 3 = 2x + 1$

2. Now, find the inverse of $f \circ g$:

   Set $y = 2x + 1$, and solve for $x$: $y = 2x + 1$ $y - 1 = 2x$ $x = \frac{y - 1}{2}$

Thus, the inverse is: $(f \circ g)^{-1}(x) = \frac{x - 1}{2}$

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4. If $f(x) = 3x - 2$ and $(f \circ g)(x) = 6x + 1$, find $g(x)$:

1. Start by expressing $(f \circ g)(x)$: $(f \circ g)(x) = f(g(x)) = 3g(x) - 2$ Given that $(f \circ g)(x) = 6x + 1$, we have: $3g(x) - 2 = 6x + 1$

2. Solve for $g(x)$:

   $3g(x) = 6x + 1 + 2 = 6x + 3$ $g(x) = \frac{6x + 3}{3} = 2x + 1$

So, $g(x) = 2x + 1$.

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5. From problem 1, find:

a) $(f \circ g)(2)$:

Recall from problem 1 that $(f \circ g)(x) = 2x^2 - 4$. Now, substitute $x = 2$: $(f \circ g)(2) = 2(2^2) - 4 = 2(4) - 4 = 8 - 4 = 4$

b) $(g \circ f)(-2)$:

Recall from problem 1 that $(g \circ f)(x) = 4x^2 + 8x + 1$. Now, substitute $x = -2$: $(g \circ f)(-2) = 4(-2)^2 + 8(-2) + 1 = 4(4) - 16 + 1 = 16 - 16 + 1 = 1$

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Let me know if you need further details or explanations!

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Related Questions:

1. Can you explain the step-by-step process of finding the inverse of a composition of functions?
2. What are the main conditions for two functions to be inverse to each other?
3. How does the domain of a composite function differ from that of individual functions?
4. What are some real-life applications of function compositions?
5. Can you provide an example where the composition of two functions results in a constant function?

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Tip:

When finding the composition of two functions, always apply the second function first and then the first function to avoid errors.