Answer
See explanations.
Work Step by Step
Step 1. To show that the function $g(t)=\frac{1}{1-t}+\sqrt {1+t}-3.1$ has exactly one zero in the interval of $(-1,1)$, we will need to show that the function crosses the x-axis only once.
Step 2. Taking the derivative of the function, we get $g'(t)=\frac{1}{(1-t)^2}+\frac{1}{2\sqrt {1+t}}$. As each term is positive, we have $g'(t)\gt0$, which means that the function $g(t)$ will only go in one direction (increases as $t$ increases).
Step 3. Letting $x=0$, we have $g(0)=\frac{1}{1-0}+\sqrt {1+0}-3.1=-1.1$, which is negative. Letting $x=0.9$, we have $g(0.9)=\frac{1}{1-0.9}+\sqrt {1+0.9}-3.1=6.9+\sqrt {1.9}\gt0$, thus the curve should cross the x-axis between $0$ and $0.9$. As a matter of fact, we can find the zero as $x=0.47$ as shown in the figure.
Step 4. Because $g'(t)\gt0$, the function has only one solution at $x=0.47$.
