The difference between the local extreme valu | vedclass

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(D) Step $1$: Find the derivative $f'(x) = 6x^2 - 30x + 36$.Step $2$: Set $f'(x) = 0$ to find critical points: $6(x^2 - 5x + 6) = 0 \implies 6(x-2)(x-

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(D) Step $1$: Find the derivative $f'(x) = 6x^2 - 30x + 36$.Step $2$: Set $f'(x) = 0$ to find critical points: $6(x^2 - 5x + 6) = 0 \implies 6(x-2)(x-3) = 0$. Thus, $x = 2$ and $x = 3$.Step $3$: Calculate the local maximum value at $x = 2$: $f(2) = 2(8) - 15(4) + 36(2) + 40 = 16 - 60 + 72 + 40 = 68$.Step $4$: Calculate the local minimum value at $x = 3$: $f(3) = 2(27) - 15(9) + 36(3) + 40 = 54 - 135 + 108 + 40 = 67$.Step $5$: The difference between the local extreme values is $|68 - 67| = 1$.

The difference between the local extreme values of the function $f(x) = 2x^3 - 15x^2 + 36x + 40$ is ......

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