From the image you provided, the question appears to involve composite functions, where $f(x)$ and $g(x)$ are given as:

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

The task is to find various composite functions and values, such as:

• $(f \circ g)(0)$
• $(g \circ f)(3)$
• $(g \circ f)(x) = 8$, solve for $x$

Step-by-Step Solution:

1. $(f \circ g)(0)$

This is the value of $f(g(0))$.

• First, find $g(0)$: $g(0) = 0^2 - 3(0) + 4 = 4$
• Now, substitute $g(0) = 4$ into $f(x)$: $f(4) = 2(4) + 1 = 8 + 1 = 9$ Thus, $(f \circ g)(0) = 9$.

2. $(g \circ f)(3)$

This is the value of $g(f(3))$.

• First, find $f(3)$: $f(3) = 2(3) + 1 = 6 + 1 = 7$
• Now, substitute $f(3) = 7$ into $g(x)$: $g(7) = 7^2 - 3(7) + 4 = 49 - 21 + 4 = 32$ Thus, $(g \circ f)(3) = 32$.

3. Solve $(g \circ f)(x) = 8$

This means we need to solve for $x$ such that $g(f(x)) = 8$.

• First, find $f(x)$: $f(x) = 2x + 1$
• Now, substitute $f(x)$ into $g(x)$: $g(f(x)) = (2x + 1)^2 - 3(2x + 1) + 4$
• Simplify the expression: $(2x + 1)^2 = 4x^2 + 4x + 1$ $-3(2x + 1) = -6x - 3$ $g(f(x)) = 4x^2 + 4x + 1 - 6x - 3 + 4 = 4x^2 - 2x + 2$
• Now, solve $4x^2 - 2x + 2 = 8$: $4x^2 - 2x + 2 - 8 = 0$ $4x^2 - 2x - 6 = 0$ Divide through by 2: $2x^2 - x - 3 = 0$
• Solve using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-3)}}{2(2)}$ $x = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm \sqrt{25}}{4}$ $x = \frac{1 \pm 5}{4}$ Thus, $x = \frac{1 + 5}{4} = \frac{6}{4} = 1.5$ or $x = \frac{1 - 5}{4} = \frac{-4}{4} = -1$.

So, the solutions are $x = 1.5$ and $x = -1$.

Would you like more details or have any questions?

Here are 5 related questions for further practice:

1. What is $(f \circ g)(x)$ in general form?
2. How would you solve $(f \circ g)(x) = 5$?
3. Can you find $(g \circ f)(0)$?
4. What are the solutions to $g(f(x)) = 16$?
5. If $f(x) = 3x - 2$ and $g(x) = x^2 + x + 1$, what is $(f \circ g)(2)$?

Tip: Always check the order of composition carefully—$f(g(x))$ is not the same as $g(f(x))$!