Answer:
1. f(x)= [tex]\left \{ {{1/xWHEN x<1 ANDx\neq 0} \atop {\sqrt[3]{x}WHEN x\geq 1}} \right.[/tex]
2. yes this graph represents a polynomial, this line is of the form y=mx+b; there are no turning points as it is a polynomial of degree 1.
Step-by-step explanation:
For problem 1:
as can be seen in the given graph, the graph of 1/x is ended at the value of x=1 and for all the values of x>1.
so the condition x<1 and x[tex]\neq[/tex]0 applies here
then for graph [tex]\sqrt[3]{x}[/tex], the graph starts from (1,1) and there is no extension of it beyond the value of x=1.
so the condition x[tex]\geq[/tex]1 applies here.
For Problem 2:
the given graph is a linear line which is represent by polynomial y=mx+b
and the polynomial y=mx+b is a polynomial of degree 1.
As the given graph is a graph of straight line represented by y=mx+b there are no turning points.
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