# Find the maximum value of f(x) = x^3 - 3x if x^4 + 36 < 13x^2. How do I do this? What are the steps.

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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

`x^4+36<13x^2` .

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`

`(x^2 -9)(x^2-4)<0`

`(x-3)(x+3)(x-2)(x+2)<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:

`3(x^2-1)=3(x-1)(x+1)=0`

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` .