Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that

$f'(x) = (x - 2024)^3(x - 2025)(x - 2026)^2$ for all $x \in \mathbb{R}$.

Let $g: \mathbb{R} \to (0, \infty)$ be a function such that $g(x) = \sqrt{f(x)}$ for all $x \in \mathbb{R}$.

Then the number of points at which $g(x)$ has a local maximum is:

Given $f'(x) = (x - 2024)^3(x - 2025)(x - 2026)^2$, we need to find local maxima of $g(x) = \sqrt{f(x)}$.

Since $g(x) = \sqrt{f(x)}$, differentiating:

$g'(x) = \frac{f'(x)}{2\sqrt{f(x)}}$

For critical points, $g'(x) = 0$ when $f'(x) = 0$, giving us $x = 2024, 2025, 2026$.

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To determine which critical points are local maxima, we analyze the sign of $f'(x)$ in each interval.

The factor $(x - 2026)^2$ is always non-negative, so the sign depends on $(x - 2024)^3(x - 2025)$:

Interval	$(x - 2024)^3$	$(x - 2025)$	$f'(x)$ sign	$f(x)$ behavior
$x < 2024$	$-$	$-$	$+$	Increasing
$2024 < x < 2025$	$+$	$-$	$-$	Decreasing
$x > 2025$	$+$	$+$	$+$	Increasing

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From the behavior of $f(x)$:

• At $x = 2024$: $f$ changes from increasing to decreasing → local maximum
• At $x = 2025$: $f$ changes from decreasing to increasing → local minimum
• At $x = 2026$: $f$ keeps increasing (no sign change) → no extremum

Since $g(x) = \sqrt{f(x)}$ is a monotonically increasing transformation of $f(x)$ (for $f > 0$), $g$ inherits the same extremum behavior as $f$.

Therefore, $g(x)$ has a local maximum only at $x = 2024$.

The answer is 1.