If the function $f(x) = 2x^3 + 9x^2+ 12x-1$ is given, then f(x) have

local minima at x = -1

The correct answer is Option (1) → local minima at x = -1

Given: $f(x) = 2x^3 + 9x^2 + 12x - 1$

First derivative: $f'(x) = 6x^2 + 18x + 12 = 6(x^2 + 3x + 2) = 6(x+1)(x+2)$

Set $f'(x) = 0$ to find critical points: $x = -1, -2$

Second derivative: $f''(x) = 12x + 18$

Check second derivative at critical points:

$f''(-1) = 12(-1)+18 = 6 > 0 \Rightarrow x=-1$ is local minimum

$f''(-2) = 12(-2)+18 = -6 < 0 \Rightarrow x=-2$ is local maximum

Evaluation of statements:

- Local minima at $x=-1$ ✔ correct

- Local maxima at $x=3/2$ ✖ incorrect

- Neither maxima nor minima at $x=-2$ ✖ incorrect ($x=-2$ is local maximum)

- Minimum value of $f(x)$ is 3/2 ✖ incorrect (minimum value at $x=-1$ is $f(-1) = 2(-1)^3 + 9(-1)^2 + 12(-1) -1 = -2$)