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(C) To find the absolute maximum and minimum values of $f(x) = 2x^3 - 15x^2 + 36x - 30$ on $[-1, 4]$,we first find the critical points by setting $f'(

Vedclass · Accepted Answer

$85$

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(C) To find the absolute maximum and minimum values of $f(x) = 2x^3 - 15x^2 + 36x - 30$ on $[-1, 4]$,we first find the critical points by setting $f'(x) = 0$.$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$.The critical points are $x = 2$ and $x = 3$,both of which lie in $[-1, 4]$.Now,we evaluate $f(x)$ at the critical points and the endpoints of the interval:$f(-1) = 2(-1)^3 - 15(-1)^2 + 36(-1) - 30 = -2 - 15 - 36 - 30 = -83$.$f(2) = 2(8) - 15(4) + 36(2) - 30 = 16 - 60 + 72 - 30 = -2$.$f(3) = 2(27) - 15(9) + 36(3) - 30 = 54 - 135 + 108 - 30 = -3$.$f(4) = 2(64) - 15(16) + 36(4) - 30 = 128 - 240 + 144 - 30 = 2$.The absolute maximum value is $2$ and the absolute minimum value is $-83$.The difference is $2 - (-83) = 85$.

The difference between the absolute maximum and absolute minimum values of the function $f(x)=2x^3-15x^2+36x-30$ on the interval $[-1, 4]$ is:

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