Given: `f(x)=5+54x-2x^3,[0,4]`

Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=54-6x^2=0`

`54=6x^2`

`9=x^2`

`x=3,x=-3`

The critical value is x=3. The value x=-3 is not in the given interval [0, 4]. Therefore x=-3 will not be used to find the absolute maximum...

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Given: `f(x)=5+54x-2x^3,[0,4]`

Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=54-6x^2=0`

`54=6x^2`

`9=x^2`

`x=3,x=-3`

The critical value is x=3. The value x=-3 is not in the given interval [0, 4]. Therefore x=-3 will not be used to find the absolute maximum or absolute minimum value. Plug in the critical value x=3 and the endpoints of the interval [0, 4] into the original f(x) function.

f(0)=5

f(3)=113

f(4)=93

Examine the f(x) value to determine the absolute maximum and absolute minimum.

The absolute maximum occurs at the point (3, 113).

The absolute minimum occurs at the point (0, 5).