Consider the following statements: Statement-I: The function f(x) =...

Given $f(x) = \dfrac{x^3 + 128}{x} = x^2 + \dfrac{128}{x}$

Differentiating with respect to $x$, we get:
$f'(x) = 2x - \dfrac{128}{x^2} = \dfrac{2(x^3 - 64)}{x^2}$

For critical points, $f'(x) = 0$
$\Rightarrow x^3 - 64 = 0 \Rightarrow x = 4$

At $x = 4$, $f(4) = \dfrac{4^3 + 128}{4} = \dfrac{192}{4} = 48$

Now, checking the sign of $f'(x)$ around $x = 4$:
For $x < 4$, $x^3 < 64 \Rightarrow f'(x) < 0$ (negative)
For $x > 4$, $x^3 > 64 \Rightarrow f'(x) > 0$ (positive)

Since $f'(x)$ changes sign from negative to positive as $x$ increases through $4$, $x = 4$ is a point of local minimum. The minimum value is $48$.

Thus, Statement-I is correct.

Statement-II claims that $f'(x)$ changes sign from positive to negative, which is incorrect.

Therefore, Statement-I is correct but Statement-II is not correct.

Answer: Statement-I is correct but Statement-II is not correct