We are tasked to find $g(x)$, given:

1. $f(x) = \frac{x^2 + 1}{2}$
2. $f(g(2-x)) = x+2$

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Step-by-Step Solution:

1. Define $f(g(2-x))$:

Substituting $f(x) = \frac{x^2 + 1}{2}$ into $f(g(2-x)) = x+2$, we get: $f(g(2-x)) = \frac{(g(2-x))^2 + 1}{2}$ Thus: $\frac{(g(2-x))^2 + 1}{2} = x + 2$

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2. Eliminate the fraction:

Multiply through by 2: $(g(2-x))^2 + 1 = 2(x+2)$ Simplify the right-hand side: $(g(2-x))^2 + 1 = 2x + 4$ Subtract 1 from both sides: $(g(2-x))^2 = 2x + 3$

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3. Solve for $g(2-x)$:

Take the square root of both sides: $g(2-x) = \pm \sqrt{2x + 3}$

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4. Express $g(x)$:

Since $g(2-x) = \pm \sqrt{2x + 3}$, let $x \to 2-y$, so $g(y) = \pm \sqrt{4 - 2y + 3}$. Simplify: $g(x) = \pm \sqrt{-2x + 7}$

Thus, the possible forms of $g(x)$ are: $g(x) = \sqrt{-2x + 7} \quad \text{or} \quad g(x) = -\sqrt{-2x + 7}.$

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Final Answer:

$g(x) = \pm \sqrt{-2x + 7}.$

Would you like me to explain further steps or verify a specific case?

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Related Questions:

1. How do you verify if $g(x)$ satisfies the given $f(g(2-x))$?
2. What happens if $g(x)$ takes only one branch of $\pm \sqrt{-2x + 7}$?
3. How do you check the domain of $g(x)$?
4. What is the behavior of $g(x)$ when $x \to \infty$?
5. How does the solution change if $f(x)$ has a different expression?

Tip: Always consider the domain of the function when dealing with square roots, as the argument under the square root must be non-negative.