# Range for the function $f(x) = 3x + 2$ with domain $x > 0$

The function below is defined for continuous domains

Sketch the graph and state the range of the function

Question: $f(x) = 3x + 2$ for the domain $\{x \in \mathbb{R} : x > 0 \}$

The straight line cuts the $y$-axis at $(2, 0)$ but since $x >0$ why is the answer for the range still $f(x) > 2$? Shouldn't you be using $x = 0.1$ or $x = 1$ to find the $y$-intercept since the domain originally stated $x > 0$ (hence $x \not= 0$)

Thanks.

Edit: I have just noticed that the range is $f(x) > 2$ and not $f(x) \ge 2$ which solves everything.

• It be $(2,\infty)$ – user60887 Apr 13 '14 at 17:18
• If $x>0$, then $f(x)=3x+2>3\cdot 0+2=2$. – J.R. Apr 13 '14 at 17:18

$x>0\Rightarrow 3x+2>2\Rightarrow f(x)\in (2,\infty)$
Yeah... But we first drew the graph of function over $R$ and then smartly erased the part not needed. Your case will yield the same answer.