How do you find the domain and range for $y = \sqrt{4 x - 1}$ $y =sqrt(4x-1)$ ?

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Answer 1 · 2015-05-25T23:05:10

Assuming we're working with real numbers, $\sqrt{a}$ $sqrt(a)$ is only defined for $a \geq 0$ $a>=0$ , but it is defined for all $a \geq 0$ $a>=0$ .

So we require $4 x - 1 \geq 0$ $4x - 1 >= 0$ .

Adding $1$ $1$ to both sides, then dividing both sides by $4$ $4$ we get:

$x \geq \frac{1}{4}$ $x >= 1/4$

So our domain is ${x in RR: x >= 1/4}$ or $[1/4, oo)$ in interval notation.

The range is ${y in RR: y >= 0}$ or $[0, oo)$ in interval notation.

To prove this:

Given any $y >= 0$ , let $x = (y^2+1)/4$

Then $4x - 1 = y^2$

and since $y >= 0$ we have $y = sqrt(4x-1)$ as required.

How do you find the domain and range for y =sqrt(4x-1)y=√4x−1y =sqrt(4x-1)?