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#Analysis# The given piecewise function is defined as follows: $$ f(x)= \begin{cases} -x & \text{if}\ x\leq 0\\ x^2 & \text{if}\ x>0 \end{cases} $$ We need to examine if this function satisfies these conditions: 1. The function is one-to-one on the interval \((-\infty, 0]\). 2. The function is not one-to-one on the interval \((-\infty, \infty)\). #Solution# #tag_title#Step 1: Verify the one-to-one property on \((-\infty, 0]\) #tag_content#On the interval \((-\infty, 0]\), the function is defined as \(f(x)=-x\). This is a linear function with a negative slope, which means that for each value of x in this interval, there will be exactly one corresponding value of y. This satisfies Condition 1, as the function is one-to-one on the interval \((-\infty, 0]\). #tag_title#Step 2: Verify the non-one-to-one property on \((-\infty, \infty)\) #tag_content#In the entire real line \((-\infty, \infty)\), the piecewise function includes both the linear segment \(-x\) for \(x\leq 0\) and the parabolic segment \(x^2\) for \(x>0\). When both of these segments are combined, the function will have multiple x-values with the same y-value. For example, \(f(-2)=(-2)^2=4\) and \(f(-1)=(-1)^2=1\) are both on the linear segment, while \(f(1)=1^2=1\) and \(f(2)=2^2=4\) are on the parabolic segment. Thus, Condition 2 is also satisfied, as the function is not one-to-one on the interval \((-\infty, \infty)\). #tag_title#Conclusion #tag_content#The given piecewise function is one-to-one on the interval \((-\infty, 0]\), but it is not one-to-one on the entire real line \((-\infty, \infty)\). The function fulfills both conditions, which makes it a suitable example for the problem statement.