which of the following is the graph for this piecewise defined function? \n$f(x) = \\begin{cases} x &…
which of the following is the graph for this piecewise defined function? \n$f(x) = \\begin{cases} x & \\text{if } x \\leq 0 \\\\ x^2 & \\text{if } x > 0 \\end{cases}$
Answer
Explanation:
Step1: Analyze ( f(x) = x ) for ( x \leq 0 )
The function ( y = x ) is a linear function with a slope of 1. For ( x \leq 0 ), we include the point at ( x = 0 ) (since the condition is ( x \leq 0 )), so the graph is a line passing through the origin with a slope of 1, to the left of and including the y - axis.
Step2: Analyze ( f(x)=x^{2} ) for ( x > 0 )
The function ( y=x^{2} ) is a parabola opening upwards. For ( x > 0 ), we do not include the point at ( x = 0 ) (since the condition is ( x>0 )), and as ( x ) increases from 0, ( y = x^{2}) increases quadratically.
Step3: Compare with the graphs
- The first graph: The left - hand side (for ( x\leq0 )) looks like ( y = |x| ) (a V - shape) instead of ( y=x ), so it is incorrect.
- The second graph: For ( x\leq0 ), it is the line ( y = x ) (including the point at ( x = 0 )), and for ( x>0 ), it is the parabola ( y=x^{2} ) (starting from the origin, but since ( x > 0 ) and the point at ( x = 0 ) is included in the ( y=x ) part, this matches our function.
- The third graph: The point at ( x = 0 ) for the ( y=x ) part is an open circle, but our function has ( f(0)=0 ) (since when ( x = 0 ), we use ( f(x)=x )), so it is incorrect.