1. Given function

We have $f(x)=2x+3x^{2/3}, \, x\in \mathbb{R}.$

We need to find the number of local maxima and local minima.

2. Find the derivative

For $x\neq 0$, $f'(x)=2+3\cdot \frac{2}{3}x^{-1/3}=2+2x^{-1/3}.$ So, $f'(x)=2+\frac{2}{x^{1/3}}, \qquad x\neq 0.$

At $x=0$, $f'(x)$ is not defined.

3. Critical points

Critical points occur where either:

• $f'(x)=0$, or
• $f'(x)$ is not defined but $f(x)$ is defined.

(i) Solve $f'(x)=0$

$2+2x^{-1/3}=0$ $1+x^{-1/3}=0$ $x^{-1/3}=-1$ $x^{1/3}=-1$ $x=-1.$

(ii) Where derivative is undefined

At $x=0$, derivative does not exist, but $f(0)$ exists.

So the critical points are: $x=-1,\; 0.$

4. Sign analysis of $f'(x)$

We study $f'(x)=2+2x^{-1/3}.$

Consider intervals:

• $(-\infty,-1)$
• $(-1,0)$
• $(0,\infty)$

On $(-\infty,-1)$

Take $x=-8$: $x^{1/3}=-2 \Rightarrow x^{-1/3}=-\frac12.$ Then $f'(-8)=2+2\left(-\frac12\right)=1>0.$ So $f'(x)>0$ on $(-\infty,-1)$.

On $(-1,0)$

Take $x=-\frac18$: $x^{1/3}=-\frac12 \Rightarrow x^{-1/3}=-2.$ Then $f'\left(-\frac18\right)=2+2(-2)=-2<0.$ So $f'(x)<0$ on $(-1,0)$.

On $(0,\infty)$

Take $x=1$: $x^{-1/3}=1,$ so $f'(1)=2+2=4>0.$ Thus $f'(x)>0$ on $(0,\infty)$.

5. Determine local extrema

At $x=-1$

$f'(x)$ changes from positive to negative: $+ \to -$ Hence, $x=-1$ is a point of local maximum.

At $x=0$

$f'(x)$ changes from negative to positive: $- \to +$ Hence, $x=0$ is a point of local minimum.

6. Conclusion

The function has:

• exactly one point of local maximum,
• exactly one point of local minimum.

So the correct option is $\boxed{\text{B}}.$

7. Comparison with stored answer

Stored correct answer: B.

Our derived answer is also B, so they agree.