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The given function f(x) is a polynomial function, so it is defined for all x. However, in this case, the domain is restricted by the given inequality
Start by finding thw intervals of x for which f(x) is defined by solving this inequality. It can be solved by factoring:
`x^4-13x^2 + 36<0`
The left side of this inequality changes sign when x = -3, -2, 2 and 3. Since for any x greater than 3, for example, x = 4, the left side is positive,
( (4-3)(4+3)(4-2)(4+2) = 84 > 0 )
then it will be negative when x is between 2 and 3, then positive again when x is between -2 and 2, then negative when x is between -3 and 2, and positive when x is less than -3. So, the solutions of the inequality above are intervals
(-3,-2), (2, 3). Notice that the ends of the intervals, x = 3, 2, -2, -3, are not included in the domain, so the function cannot be maximum there.
To determine where the given function will have a maximum, we need to find its derivative:
The derivative of f(x) is
`f'(x) = 3x^2- 3`
It will be zero when `3x^2-3=0` . This quadratic equation can be solved by factoring:
The roots of this equation are x = 1 and x = -1. So the function f(x) would be maximum at one of these points, but these points do not lie inside the domain.
We can see that the derivative is positive when x is greater than 1 and less than -1. This means that the function is increasing on these intervals. Notice also that these intervals include the domain of the function, intervals
(2, 3) and (-3, -2).
Since the function is only increasing on these intervals, and the ends of the intervals are not included in the domain, the given function cannot attain a maximum value on this domain.
If the ends were included, then the function would be maximum at x =3:
`f(3) = 3^3-3*3=18` .
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