You need to remember the absolute value property such that:

`|x|<c => -c < x < c`

Reasoning by analogy yields:

`(1/3)x^2 - 3<= 3 - |x| <= 3 - (1/3)x^2`

Subtracting 3 both sides yields:

`(1/3)x^2 - 6<= -|x| <= - (1/3)x^2`

Multiplying by -1 yields:

`6 - (1/3)x^2 >=...

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You need to remember the absolute value property such that:

`|x|<c => -c < x < c`

Reasoning by analogy yields:

`(1/3)x^2 - 3<= 3 - |x| <= 3 - (1/3)x^2`

Subtracting 3 both sides yields:

`(1/3)x^2 - 6<= -|x| <= - (1/3)x^2`

Multiplying by -1 yields:

`6 - (1/3)x^2 >= |x| >= (1/3)x^2`

You should solve the inequality `|x| <= 6 - (1/3)x^2 ` such that:

`|x| <= 6 - (1/3)x^2 => (1/3)x^2 - 6 <= x <= 6 - (1/3)x^2`

Solving the inequality `(1/3)x^2 - 6 <= x` yields:

`(1/3)x^2 - x- 6 <= 0`

You need to solve the equation `(1/3)x^2 - x - 6 = 0 ` such that:

`x^2 - 3x - 18 = 0`

Using quadratic formula yields:

`x_(1,2) = (3+-sqrt(9 + 72))/2 => x_(1,2) = (3+-sqrt81)/2`

`x_1 = 6 ; x_2 = -3`

Notice that the inequality holds for `x in [-3 , 6].`

Solving the inequality `x <= 6 - (1/3)x^2` yields:

`(1/3)x^2 + x - 6 <= 0`

`x^2 + 3x - 18 = 0 => x_1 = 6 ; x_2 = -3 => x in [-3 , 6]`

You should solve the inequality `|x| >= (1/3)x^2` such that:

`-(1/3)x^2 >= x >= (1/3)x^2`

`(1/3)x^2 - x <= 0`

`x(x/3 - 1) <= 0`

You need to solve for x the equation `x(x/3 - 1) = 0` such that:

`x = 0`

`x/3 - 1 = 0 => x - 3 = 0 => x = 3`

Notice that the inequality holds for `x in [0,3].`

Hence, you need to find the common interval of values of x for the inequality holds such that:

`x in [-3 , 6] nn [0 , 3] => x in [0 , 3]`

**Hence, evaluating the interval of values of x for the inequality holds yields `x in [0 , 3].` **