To find the region where the function $f(x) = -\frac{(x-1)^2}{2} + 3$ is decreasing, we need to analyze the first derivative of the function and find where it is negative.

Step 1: Find the first derivative of $f(x)$.

The given function is:

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

Now, we calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx} \left[ -\frac{(x-1)^2}{2} + 3 \right]$

Using the chain rule, we get:

$f'(x) = -\frac{2(x-1)}{2} = -(x-1)$

Step 2: Analyze the sign of $f'(x)$.

The derivative is $f'(x) = -(x-1)$. To determine where $f(x)$ is decreasing, we need to find where $f'(x) < 0$:

$-(x-1) < 0$

Solving this inequality:

$x-1 > 0 \quad \Rightarrow \quad x > 1$

Conclusion:

The function $f(x)$ is decreasing for $x > 1$.

Would you like further details or have any questions?

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Here are 5 related questions:

1. What is the region where $f(x)$ is increasing?
2. How do we find critical points of a function?
3. What is the maximum or minimum value of $f(x)$?
4. How do second derivatives help in determining concavity?
5. Can you sketch the graph of $f(x)$?

Tip: The sign of the first derivative gives important information about where a function is increasing or decreasing!