Question

Determine the range of the function $f(x) = \frac{2025^{\tan x}}{45^{\tan x}+1}$.

Answer 1

(0, \infty)

Answer 2

To determine the range of the function $f(x) = \frac{2025^{\tan x}}{45^{\tan x}+1}$, we can follow these steps:

1. Simplify the expression: Notice that $2025 = 45^2$. Substitute this into the function: $f(x) = \frac{(45^2)^{\tan x}}{45^{\tan x}+1} = \frac{(45^{\tan x})^2}{45^{\tan x}+1}$

2. Introduce a substitution: Let $u = 45^{\tan x}$. The function then becomes a function of $u$: $g(u) = \frac{u^2}{u+1}$

3. Determine the domain of the substituted variable $u$: The domain of $\tan x$ is all real numbers, $\mathbb{R}$. Since the base $45 > 0$, the exponential function $45^y$ always produces positive values. Therefore, the range of $u = 45^{\tan x}$ is $(0, \infty)$. We need to find the range of $g(u)$ for $u \in (0, \infty)$.

4. Analyze the function $g(u)$ for its range: We can analyze the behavior of $g(u)$ by considering its derivative: $g'(u) = \frac{d}{du}\left(\frac{u^2}{u+1}\right)$ Using the quotient rule, $g'(u) = \frac{(2u)(u+1) - (u^2)(1)}{(u+1)^2} = \frac{2u^2 + 2u - u^2}{(u+1)^2} = \frac{u^2 + 2u}{(u+1)^2} = \frac{u(u+2)}{(u+1)^2}$.

   For $u \in (0, \infty)$:

   • $u > 0$
   • $u+2 > 0$
   • $(u+1)^2 > 0$ Therefore, $g'(u) > 0$ for all $u \in (0, \infty)$. This indicates that $g(u)$ is strictly increasing on its domain $(0, \infty)$.

   Now, we evaluate the limits of $g(u)$ as $u$ approaches the boundaries of its domain:

   • As $u \to 0^+$: $\lim_{u \to 0^+} g(u) = \lim_{u \to 0^+} \frac{u^2}{u+1} = \frac{0^2}{0+1} = 0$
   • As $u \to \infty$: $\lim_{u \to \infty} g(u) = \lim_{u \to \infty} \frac{u^2}{u+1}$ We can divide the numerator and denominator by $u$: $\lim_{u \to \infty} \frac{u}{1 + \frac{1}{u}} = \frac{\infty}{1+0} = \infty$

   Since $g(u)$ is strictly increasing on $(0, \infty)$ and its limits at the boundaries are $0$ and $\infty$, the range of $g(u)$ for $u \in (0, \infty)$ is $(0, \infty)$.

Therefore, the range of the function $f(x)$ is $(0, \infty)$.