For the function $f(x) = 2x^3 - 3x^2 - 12x + 5$, the difference of maximum and minimum value of $f(x)$ is

For the function $f(x) = 2x^3 - 3x^2 - 12x + 5$, find the derivative:

$f'(x) = 6x^2 - 6x - 12$

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To find critical points, set $f'(x) = 0$:

$6x^2 - 6x - 12 = 0$

$x^2 - x - 2 = 0$

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

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

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Use the second derivative test to classify the critical points:

$f''(x) = 12x - 6$

At $x = -1$:

$f''(-1) = 12(-1) - 6 = -18$

Since $f''(-1) < 0$, $x = -1$ gives a maximum.

At $x = 2$:

$f''(2) = 12(2) - 6 = 18$

Since $f''(2) > 0$, $x = 2$ gives a minimum.

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Maximum value at $x = -1$:

$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5$

$f(-1) = -2 - 3 + 12 + 5$

$f(-1) = 12$

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Minimum value at $x = 2$:

$f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 5$

$f(2) = 16 - 12 - 24 + 5$

$f(2) = -15$

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The difference between maximum and minimum values:

$12 - (-15) = 12 + 15 = 27$

Therefore, the difference is $27$.