Question

Please explain your answer as well. THX!!!!!

Answer 1

1. f(x)=
`\left \{ {{1/xWHEN x<1 ANDx\\eq 0} \atop {\sqrt[3]{x}WHEN x\geq 1}} \right.`

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.

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
`\\eq`0 applies here

then for graph
`\sqrt[3]{x}`, the graph starts from (1,1) and there is no extension of it beyond the value of x=1.

so the condition x
`\geq`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.

!