Question

Determine whether each relation defines y as a function of x. Give the domain and range. See Example 5. y = √(4x + 1)

Accepted Answer

Step 1: Understand the given relation. The equation is $$y = \sqrt{4x + 1}. $$This means $$y is $$defined as the square root of the expression $$4x + 1. $$Step 2: Determine if $$y is a $$function of $$x. $$For each value of $$x$$, check if there is exactly one corresponding value of $$y. $$Since the square root function outputs only the non-negative root, for each $$x$$ that makes $$4x + 1$$ non-negative, there is exactly one $$y. $$Therefore, $$y is a $$function of $$x. $$Step 3: Find the domain of the function. The expression inside the square root, $$4x + 1$$, must be greater than or equal to zero to keep $$y$$ real. Set up the inequality: $$4x + 1 \geq 0. $$Step 4: Solve the inequality for $$x. $$Subtract 1 from both sides: $$4x \geq -1$$, then divide both sides by 4: $$x \geq -\frac{1}{4}. So$$, the domain is all real numbers $$x$$ such that $$x \geq -\frac{1}{4}. $$Step 5: Determine the range of the function. Since $$y = \sqrt{4x + 1}$$ and the square root function outputs values greater than or equal to zero, the smallest value of $$y is 0 ($$when $$x = -\frac{1}{4}). As$$ $$x$$ increases, $$y$$ increases without bound. Therefore, the range is $$y \geq 0.$$

Determine whether each relation defines y as a function of x. Give the domain and range. See Example 5. y = √(4x + 1)

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Concetti chiave

Definition of a Function

Domain of a Function

Range of a Function