The function $$f(x) = 2x + 3x^{\frac{2}{3}}, x \in R$$, has

We need to determine the local maxima and minima of $$f(x) = 2x + 3x^{2/3}$$ for $$x \in \mathbb{R}$$.

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

This can be written as:

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

Critical points occur where $$f'(x) = 0$$ or $$f'(x)$$ is undefined.

Case 1: $$f'(x) = 0$$: The numerator $$2(x^{1/3} + 1) = 0 \implies x^{1/3} = -1 \implies x = -1$$.

Case 2: $$f'(x)$$ undefined: The denominator $$x^{1/3} = 0 \implies x = 0$$.

So the critical points are $$x = -1$$ and $$x = 0$$.

At $$x = -1$$: $$f'(x)$$ changes from $$+ve$$ to $$-ve$$ → local maximum.

$$f(-1) = 2(-1) + 3(-1)^{2/3} = -2 + 3(1) = 1$$

At $$x = 0$$: $$f'(x)$$ changes from $$-ve$$ to $$+ve$$ → local minimum.

$$f(0) = 0$$

Conclusion: The function has exactly one point of local maxima (at $$x = -1$$) and exactly one point of local minima (at $$x = 0$$).

The answer is Option C: exactly one local maxima and exactly one local minima.