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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.

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

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  • $\begingroup$ I have just noticed that the range is f(x) > 2 and not f(x) =>2 which solves everything. $\endgroup$ – Noah Apr 13 '14 at 17:24
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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.

Your Ad Here answers the range doubt.

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  • $\begingroup$ Why was this downvoted? $\endgroup$ – evil999man Apr 13 '14 at 17:57

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