# Given -2<=x=<5 and f(x)=x^2, what is wrong in (-2)^2=<x^2=<5^2?(-2)^2=<x^2=<5^2 is my solution and is wrong. what is wrong!!?? thanks!

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### 1 Answer

The problem provides the information that `x in [-2;5]` and `f(x) = x^2` , hence, you should determine the monotony of function over [-2;0] and [0;5] such that:

`x = -2 =gt f(-2) = (-2)^2 = 4`

`x = 0 =gt f(0) = 0`

Notice that `-2 lt 0` but `f(-2) = 4 gt f(0) = 0` , hence, the function decreases over `[-2;0] ` and you need to change the sense of inequality for `x in [-2;0]` such that:

`(-2)^2 gt= x^2 if x in [-2;0]`

You need to analyze the monotony of function for `x in [0;5]` such that:

`x = 0 =gt f(0) = 0 `

`x = 5 =gt f(5) = 25`

Notice that `0 lt 5 =gt f(0) = 0 lt f(5) = 25` , hence, the function increases over [0;5] and the statement `x^2 =lt 5^2` is true for `x in [0;5].`

**Hence, the function decreases over `[-2;0]` and it increases over `[0;5], ` hence `(-2)^2gt= x^2` for all `x in [-2;0]` and `x^2 =lt 25` for all `x in [0;5].` **

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