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I have

$f(x)=\sqrt{3x}+1$

$g(x)=x+1$

My thinking was that at the intersection points both will be equal to each other so

$\sqrt{3x}+1=x+1$

$\sqrt{3x}=x$

However I don't know where to go from here.

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  • $\begingroup$ tried squaring both sides? $\endgroup$ – user87543 Nov 1 '13 at 13:37
  • $\begingroup$ Ah yeah of course. Cheers $\endgroup$ – user88720 Nov 1 '13 at 13:40
  • $\begingroup$ Only thing is though, what would the intersection point be? $\endgroup$ – user88720 Nov 1 '13 at 13:42
  • $\begingroup$ what would be the value of $x$ if you solve for $3\sqrt{x}=x$? $\endgroup$ – user87543 Nov 1 '13 at 13:43
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$$\sqrt{3x} = x \implies 3x = x^2 \iff x^2 - 3x = x(x - 3) = 0$$ $$\implies x = 0 \quad \text{OR}\quad x = 3$$

Hence, the intersection points will be $(0, f(0)),$ and $(3, f(3))$.

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  • $\begingroup$ Clean and clear +1 $\endgroup$ – Amzoti Nov 2 '13 at 0:52
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what would be the value of x if you solve for $3\sqrt{x}=x$?

If you know what would be value of $x$ in suhc case then you can just find what is $f(x)$ for corresponding $x$ and set $(x,f(x))$.

That would be " an" intersection point.

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