Answer
The equation has 1 solution: $(3.516)$.
Work Step by Step
$$\sqrt x+\sqrt{1+x}=4$$ $$\sqrt x+\sqrt{1+x}-4=0$$
To prove the equation has a solution using the Immediate Value Theorem, we need to find 2 values $x_1$ and $x_2$ and show that there is a change of sign in the function $f(x)=\sqrt x+\sqrt{1+x}-4$ as $x$ goes from $x_1$ to $x_2$.
- Take $x_1=0$: $f(0)=\sqrt0+\sqrt{1+0}-4=0+1-4=-3\lt0$
- Take $x_2=4$: $f(4)=\sqrt4+\sqrt{1+4}-4=2+\sqrt5-4=\sqrt5-2\gt0$
So there is a change of sign in $f(x)$ as $x$ goes from $0$ to $4$.
- On $[0,4]$: $\lim_{x\to c}f(x)=\lim_{x\to c}(\sqrt x+\sqrt{1+x}-4)=\sqrt c+\sqrt{1+c}-4=f(c)$
So $f(x)$ is continuous on $[0,4]$.
Therefore, according to the Intermediate Value Theorem, there must be a value $x=c\in[0,4]$ such that $f(c)=0$. In other words, the equation $\sqrt x+\sqrt{1+x}-4=0$ or $\sqrt x+\sqrt{1+x}=4$ has at least one solution in the interval $[0,4]$.
The graph of the function $f(x)=\sqrt x+\sqrt{1+x}-4$ is enclosed below. Looking at the graph, it turns out that the curve $f(x)$ crosses the line $y=0$ at only $1$ point. The $x$-coordinate of that point is the solution of the equation.
So the equation has 1 solution: $(3.516)$.
