10.4. a) Investigate the function $y=\frac{\sqrt{x^{2}+1}+x-1}{\sqrt{x^{2}+1}+x+1}$ for evenness (oddness). b) Find the domain and range of this function.
Answer: a) the function is odd; b) domain of definition ( $-\infty, \infty$ ), range of values $(-1 ; 1)$. Solution. Since $\sqrt{x^{2}+1}>|x|$, then $\sqrt{x^{2}+1}+x+1>0$ for all $x$. Therefore, the domain of definition is ( $-\infty, \infty$ ). We transform the expression for $y$ by multiplying the numerator and denominator by the factor $\sqrt{x^{2}+1}-(x+1)$. Note that this factor is zero only at $x=0$ (this follows from $\sqrt{x^{2}+1}=x+1 \Leftrightarrow x^{2}+1=x^{2}+2 x+1$ ). After multiplying, the denominator will be $-2 x$, and the numerator $2\left(1-\sqrt{x^{2}+1}\right)$ (in both cases, the formula for the difference of squares is used). Thus, for $x \neq 0$ we have $y=\frac{\sqrt{x^{2}+1}-1}{x}$. It is clear that this function is odd (for $x=0$, the original expression for $y$ gives the value $y=0$ ). Therefore, it is sufficient to consider $x>0$. The range of values can be found by investigating the given function using the derivative. However, we can do without the derivative as follows. The range of $y$ is the set of parameters $t$ for which the equation $\frac{\sqrt{x^{2}+1}-1}{x}=t$ has a solution (note that $t>0$ for $x>0$ ). We have $\sqrt{x^{2}+1}=t x+1 \Leftrightarrow x^{2}+1=t^{2} x^{2}+2 t x+1 \Leftrightarrow x=\frac{2 t}{1-t^{2}}$. Since we are considering $x>0$, then $0<t<1$. Taking into account the oddness of the function, we obtain the range of values $(-1 ; 1)$.