Zeros and the sign of the function

Question

I need help solving this task, if anyone had a similar problem it would help me.

The task is: Determine the zero and sign of the function $$ y=1-x+\sqrt{{x^3\over x+3}} $$

I found that the function intersects $x$ axis in $\frac{5+\sqrt{13}}{2}$ and $y$ in $1$. Is that right ? How to find sign of the function ?

Thanks in advance!

Answer 1

Well, if $y=0$ then $$ \begin{split} x - 1 &= \sqrt{\frac{x^3}{x+3}} \\ x^{3/2} &= (x-1)\sqrt{x+3} \\ x^3 &= (x-1)^2(x+3) = x^3+x^2-5x + 3 \\ 0 &= x^2-5x + 3 \\ \end{split} $$ and the quadratic formula applies to give you two solutions.

At $x = 0$, your function is positive, and when you cross a zero (in either direction), it will change sign.

Can you finish?