Answer
See explanations.
Work Step by Step
Step 1. To show that the function $g(t)=\sqrt t+\sqrt {1+t}-4$ has exactly one zero in the interval of $(0,\infty)$, we will need to show that the function crosses the x-axis only once.
Step 2. Taking the derivative of the function, we get $g'(t)=\frac{1}{2\sqrt t}+\frac{1}{2\sqrt {t+1}}$. As $t\gt0$, we have $g'(t)\gt0$, which means that the curve will only go in one direction (increases as $x$ increases).
Step 3. Letting $x=1$, we have $g(1)=\sqrt 1+\sqrt {1+1}-4=\sqrt 2-3\lt0$, which is negative. Letting $x=15$, we have $g(15)=\sqrt 15+\sqrt {1+15}-4=\sqrt {15}\gt0$; thus the curve should cross the x-axis between $1$ and $15$. As a matter of fact, we can find the zero as $x\approx3.52$ as shown in the figure.
Step 4. Because $g'(t)\gt0$, the function has only one solution at $x\approx3.52$.
