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 :

$$f'(x) = (x - 2024)^3 (x - 2025)(x - 2026)^2$$

$$g(x) = \sqrt{f(x)}$$

Differentiating both sides with respect to $$x$$, we get,

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

$$g'(x) = 0$$ when $$f'(x) = 0$$

$$\Rightarrow$$ Critical points : $$x = 2024, 2025, 2026$$

The factor $$(x - 2026)^2$$ is always positive, so the sign of $$f'(x)$$ depends on $$(x - 2024)^3 (x - 2025)$$.

For $$ x < 2024$$ $$\Rightarrow$$ $$f'(x)$$ is positive $$\Rightarrow$$ $$f(x)$$ is increasing 

For $$ 2024 < x < 2025$$ $$\Rightarrow$$ $$f'(x)$$ is negative $$\Rightarrow$$ $$f(x)$$ is decreasing

For $$ x > 2025$$ $$\Rightarrow$$ $$f'(x)$$ is positive $$\Rightarrow$$ $$f(x)$$ is increasing

Hence,

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 maximum or minimum

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

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

The answer is 1.