How do you find domain and range for $f (x) = \sqrt{7 x + 2}$ $f(x) = sqrt(7x + 2)$ ?

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Answer 1 · 2018-06-28T16:12:57

The domain is $x \in [- \frac{2}{7}, + \infty)$ $x in [-2/7, +oo)$ . The range is $y \in [0, + \infty)$ $y in [0,+oo)$

Let $y = \sqrt{7 x + 2}$ $y=sqrt(7x+2)$

What's under the square root sign is $\geq 0$ $>=0$

Therefore,

$7 x + 2 \geq 0$ $7x+2>=0$

$\Rightarrow$ $=>$ , $x \geq - \frac{2}{7}$ $x>=-2/7$

The domain is $x \in [- \frac{2}{7}, + \infty)$ $x in [-2/7, +oo)$

When,

$x = - \frac{2}{7}$ $x=-2/7$ , $\Rightarrow$ $=>$ , $y = 0$ $y=0$

And

$lim_(x->+oo)sqrt(7x+2)=+oo$

Therefore,

The range is $y in [0,+oo)$

graph{sqrt(7x+2) [-7.06, 21.42, -7.46, 6.78]}

How do you find domain and range for f(x) = sqrt(7x + 2)f(x)=√7x+2 f(x) = sqrt(7x + 2)?