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

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

To find critical points, take the first derivative and set it equal to zero:

$f'(x) = 6x^2 + 18x + 12$

Setting $f'(x) = 0$:

$6x^2 + 18x + 12 = 0$

$x^2 + 3x + 2 = 0$

$(x + 1)(x + 2) = 0$

Critical points: $x = -1$ and $x = -2$

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To determine the nature of each critical point, use the second derivative test:

$f''(x) = 12x + 18$

When $f''(x) > 0$, the function has a local minimum.

When $f''(x) < 0$, the function has a local maximum.

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At $x = -1$:

$f''(-1) = 12(-1) + 18$

$f''(-1) = -12 + 18$

$f''(-1) = 6 > 0$

Therefore, $f(x)$ has a local minimum at $x = -1$.

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At $x = -2$:

$f''(-2) = 12(-2) + 18$

$f''(-2) = -24 + 18$

$f''(-2) = -6 < 0$

Therefore, $f(x)$ has a local maximum at $x = -2$.

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The function $f(x) = 2x^3 + 9x^2 + 12x - 1$ has a local minima at $x = -1$ and a local maxima at $x = -2$.