The absolute maximum value of the function f( | vedclass

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(C) Given the function $f(x)=2 x^3-9 x^2+12 x+1$ on the interval $[0,2]$.First, find the critical points by calculating the derivative $f'(x)$ and set

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Given the function $f(x)=2 x^3-9 x^2+12 x+1$ on the interval $[0,2]$.First, find the critical points by calculating the derivative $f'(x)$ and setting it to $0$:$f'(x) = 6x^2 - 18x + 12$Setting $f'(x) = 0$ gives:$6(x^2 - 3x + 2) = 0$$6(x-1)(x-2) = 0$Thus, the critical points are $x=1$ and $x=2$.Now, evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[0,2]$:At $x=0$: $f(0) = 2(0)^3 - 9(0)^2 + 12(0) + 1 = 1$At $x=1$: $f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 1 = 2 - 9 + 12 + 1 = 6$At $x=2$: $f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 1 = 16 - 36 + 24 + 1 = 5$Comparing the values ${1, 6, 5}$, the absolute maximum value is $6$.

The absolute maximum value of the function $f(x)=2 x^3-9 x^2+12 x+1$ on the interval $[0,2]$ is:

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