Find count of this functions roots:$\sqrt{x+1}-x^2+1=0$

Question

There is an equation here: $$\sqrt{x+1}-x^2+1=0$$ Now we want to write the equation $f(x)$ like $h(x)=g(x)$ in a way that we know how to draw h and g functions diagram. Then we draw the h and g function diagrams and find the common points of them. So it will be number of the $f(x)$ roots that here is the equation mentioned top. Actually now my problem is with drawing the first equation's diagram I want you to draw its diagrams like $\sqrt{x-1}$ syep by step. Please help me with it!

Answer 1

Hint: Write your equation in the form $$\sqrt{x+1}=(x+1)(x-1)$$

Answer 2

Guide:

• First draw $\sqrt{x}$.

• Now think of having drawn $h(x)$, how would you draw $h(x\color{red}+1)$.

Answer 3

By the Sonnhard's hint we obtain the domain: $\{-1\}\cup[1,+\infty)$.

$-1$ is a root and after squaring of the both sides we obtain: $$1=(x+1)(x-1)^2$$ or $$x(x^2-x-1)=0,$$ which gives also $$x=\frac{1+\sqrt5}{2}.$$

Answer 4

To draw the figure, the idea is turning the equation to $\sqrt{x+1}=x^2-1$. I guess $y=x^2-1$ is easy to draw. You have an open upward parabola with $y$-intercept at $-1$ and $x$-intercepts $\pm1$.

For the square root function, you know $g(x):=\sqrt{x}$ is increasing and concave so $g$ is keep going up and flatter as $x$ rises. Moreover, it should strictly above $x$-axis for $x>0$, and has $x$-intercept $0$. Now, $\sqrt{x+1}$ is exactly shifting the whole square root curve 1 unit left.

Now counting the intersection would give you the answer. (You should see that both curves cut at $x$-axis already, and there is one more solution on $x>0$.)