$\begingroup$

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.

$\endgroup$ 4

2 Answers

$\begingroup$

$$\sqrt{3x} = x \implies 3x = x^2 \iff x^2 - 3x = x(x - 3) = 0$$ $$\implies x = 0 \quad \text{OR}\quad x = 3$$

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

$\endgroup$ 1 $\begingroup$

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.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy