JEE Main 2024 — 29 January, Shift 2 — Question 5

The function given is $f(x) = 2x + 3(x)^{2/3}$, for $x \in \mathbb{R}$. We need to determine its properties regarding extrema.

1. Find the first derivative $f'(x)$: The derivative of $f(x)$ is: $f'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(3x^{2/3})$ $f'(x) = 2 + 3 \cdot \frac{2}{3} x^{(2/3)-1}$ $f'(x) = 2 + 2x^{-1/3}$ $f'(x) = 2 + \frac{2}{x^{1/3}}$

2. Find critical points: Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Set $f'(x) = 0$: $2 + \frac{2}{x^{1/3}} = 0$ $\frac{2}{x^{1/3}} = -2$ $x^{1/3} = -1$ Cubing both sides: $x = (-1)^3$ $x = -1$.

$f'(x)$ is undefined when $x^{1/3} = 0$, which means $x = 0$. So, the critical points are $x = -1$ and $x = 0$.

3. Use the first derivative test to determine the nature of the critical points: We will check the sign of $f'(x)$ in intervals around the critical points. Consider the intervals: $(-\infty, -1)$, $(-1, 0)$, and $(0, \infty)$.

Interval $(-\infty, -1)$: Choose $x = -8$. $f'(-8) = 2 + \frac{2}{(-8)^{1/3}} = 2 + \frac{2}{-2} = 2 - 1 = 1$. Since $f'(-8) > 0$, $f(x)$ is increasing in this interval.

Interval $(-1, 0)$: Choose $x = -1/8$. $f'(-1/8) = 2 + \frac{2}{(-1/8)^{1/3}} = 2 + \frac{2}{-1/2} = 2 - 4 = -2$. Since $f'(-1/8) < 0$, $f(x)$ is decreasing in this interval. At $x = -1$, the function changes from increasing to decreasing, so there is a \textbf{local maximum} at $x = -1$. The value of the local maximum is $f(-1) = 2(-1) + 3(-1)^{2/3} = -2 + 3(1) = 1$.

Interval $(0, \infty)$: Choose $x = 1/8$. $f'(1/8) = 2 + \frac{2}{(1/8)^{1/3}} = 2 + \frac{2}{1/2} = 2 + 4 = 6$. Since $f'(1/8) > 0$, $f(x)$ is increasing in this interval. At $x = 0$, the function changes from decreasing to increasing, so there is a \textbf{local minimum} at $x = 0$. The value of the local minimum is $f(0) = 2(0) + 3(0)^{2/3} = 0 + 0 = 0$.

Conclusion: The function $f(x)$ has a local maximum at $x=-1$ and a local minimum at $x=0$. Therefore, it has both a local maximum and a local minimum.

The final answer is a local maximum and a local minimum.