Question

Determine whether each relation defines y as a function of x. Give the domain and range. y=√(4x+1)

Accepted Answer

Identify the given relation: $$y = \sqrt{4x + 1}. $$This means $$y is $$defined as the square root of the expression $$4x + 1. $$Determine the domain by finding all values of $$x$$ for which the expression inside the square root is non-negative, since the square root of a negative number is not a real number. Set up the inequality: $$4x + 1 \geq 0. $$Solve the inequality $$4x + 1 \geq 0 to $$find the domain. Subtract 1 from both sides: $$4x \geq -1$$, then divide both sides by 4: $$x \geq -\frac{1}{4}. $$Check if the relation defines $$y as a $$function of $$x. $$Since for each $$x in $$the domain there is exactly one non-negative value of $$y ($$the principal square root), this relation does define $$y as a $$function of $$x. $$Determine the range by considering the possible values of $$y. $$Since $$y = \sqrt{4x + 1}$$ and the square root function outputs only non-negative values, the smallest value of $$y is 0 ($$when $$x = -\frac{1}{4}$$), and $$y$$ increases without bound as $$x$$ increases. So, the range is $$y \geq 0.$$

Determine whether each relation defines y as a function of x. Give the domain and range. y=√(4x+1)

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Definition of a Function

Domain of a Function

Range of a Function