Find the interval on which is continuous. specify right or left continuity at the endpoints.

is continuous on its domain of .

is continuous from the left at -1 and continuous from the right at 1.

assuming domain and codomain of all real numbers, function is only defined for nonnegative radicands:

there are many ways of solving a quadratic inequality. here we will reason by cases, using the definition of absolute value.

if then and so the inequality turns into when x ≥ 0. of course, "x≥3 when x≥0" simplifies to x ≥ 3.

or, if , then and so when . of course, "x≤-3 when x < 0" simplifies to x ≤ -3.

so the domain of is

now, is a continuous function since it is a composition of continuous functions. is a polynomial and (the parent) is a root function; both of them are continuous on their domains. therefore, is continuous on its domain of .

by definition, a function is continuous from the left at a number a if and continuous from the right at a number a if .

checking continuity at the end points, note that

and

so is continuous from the left at -1 and continuous from the right at 1.

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.

1 + (6 / x - 1)

step-by-step explanation:

divide using polynomial division.