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Chapter 5: Problem 15

Examine which open [respectively, closed] intervals are mapped by \(f(x):=x^{2}\) for \(x \in \mathbb{R}\) onto open [respectively, closed] intervals.

Short Answer

Expert verified
In summary, any open interval \(-a<x<a\) is mapped onto the open interval [0, \(a^2\)), while any closed interval \(-a \leq x \leq a\) is mapped onto the closed interval [0, \(a^2\)].

Step by step solution

01

Understand the Function

First, one should acknowledge the nature of the function \(f(x)=x^2\). For every real number \(x\), the output of the function will always be a non-negative value.
02

Define the Open Interval

An open interval does not include its endpoints. So, an open interval for \(x\) could be \(-a < x < a\) where \(a\) is a positive constant.
03

Evaluate Open Intervals under function

By substituting such \(x\) into the function \(f(x)=x^2\), we find that this maps to an interval [0, \(a^2\)). The reason why the interval does not include \(a^2\) is because we can make \(x\) as close to \(a\) or \(-a\) as we want, but never equal.
04

Define the Closed Interval

A closed interval includes its endpoints. So, a closed interval for \(x\) could be \(-a \leq x \leq a\) where \(a\) is a positive constant.
05

Evaluate Closed Intervals under function

By substituting such \(x\) into function \(f(x)=x^2\), we find that this maps to an interval [0, \(a^2\)]. The reason why the interval includes \(a^2\) is because we can allow \(x\) to actually be \(a\) or \(-a\).

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