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(B) Given the function $f(x) = 2x^3 - 15x^2 + 36x - 8$. To find the turning points, we calculate the derivative $f'(x)$ and set it to $0$: $f'(x) = 6x

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$-2$

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(B) Given the function $f(x) = 2x^3 - 15x^2 + 36x - 8$. To find the turning points, we calculate the derivative $f'(x)$ and set it to $0$: $f'(x) = 6x^2 - 30x + 36 = 0$. Dividing by $6$, we get $x^2 - 5x + 6 = 0$. Factoring the quadratic equation: $(x - 2)(x - 3) = 0$. Thus, the $x$-coordinates of the turning points are $x = 2$ and $x = 3$. For $x = 2$, $f(2) = 2(8) - 15(4) + 36(2) - 8 = 16 - 60 + 72 - 8 = 20$. For $x = 3$, $f(3) = 2(27) - 15(9) + 36(3) - 8 = 54 - 135 + 108 - 8 = 19$. Given $\alpha Correspondingly, $\beta = 20$ and $\delta = 19$. Now, calculate $\alpha - \gamma - \beta + \delta = 2 - 3 - 20 + 19 = -2$.

If $(\alpha, \beta)$ and $(\gamma, \delta)$ where $\alpha < \gamma$ are the turning points of $f(x) = 2x^3 - 15x^2 + 36x - 8$, then $\alpha - \gamma - \beta + \delta =$

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