[tex]2x+1[/tex] is a line that stretches from [tex]-\infty[/tex] as [tex]x\to-\infty[/tex], to [tex]+\infty[/tex] as [tex]x\to\infty[/tex], hence with range [tex](-\infty,\infty)[/tex].
But in the context of [tex]h(x)[/tex], the function is equal to [tex]2x+1[/tex] only for those [tex]x[/tex] that are less than 1, so that this piece of [tex]h[/tex] only stretches as far as [tex]2\cdot1+1=3[/tex], which means this piece has a range of [tex](-\infty,3)[/tex].
Then for all [tex]x[/tex] beyond [tex]x=1[/tex], the function stays fixed at [tex]h(x)=3[/tex]. So the range is [tex](-\infty,3][/tex] (with 3 included in the range).
