Animated Solution for Mathematics - Functions: Let $f:(-\pi /2,\pi /2)\to R$ be given by $f(x)=(\log (\sec x+\tan x){)}^3$. Then

We are given the function $f(x)=(\log (\sec x+\tan x){)}^3$ defined on the open interval $(-\frac{\pi }{2},\frac{\pi }{2})$. Our goal is to uncover its properties regarding symmetry, injectivity, and surjectivity.

To determine if the function is odd or even, we evaluate $f(-x)$ by replacing $x$ with $-x$:

Recall the trigonometric foundations: $\sec x$ is an even function ($\sec (-x)=\sec x$), while $\tan x$ is an odd function ($\tan (-x)=-\tan x$). Substituting these, we get:

Using the identity ${\sec }^{2}x-{\tan }^{2}x=1$, we observe that $(\sec x-\tan x)(\sec x+\tan x)=1$. This implies $\sec x-\tan x=\frac{1}{\sec x+\tan x}$. Substituting this into our expression:

Applying the logarithmic property $\log (\frac{1}{A})=-\log A$, we find:

Since $f(-x)=-f(x)$, we have proven that the function is odd.

To investigate if the function is one-one, we examine its monotonicity by finding the derivative $f^{\prime }(x)$ using the chain rule:

Factoring $\sec x$ from the term $(\sec x\tan x+{\sec }^{2}x)$, we get $\sec x(\tan x+\sec x)$. The $(\sec x+\tan x)$ terms cancel out, simplifying the derivative to:

In the domain $(-\frac{\pi }{2},\frac{\pi }{2})$, $\sec x>0$ and the squared term is non-negative. Thus, $f^{\prime }(x)\ge 0$, and since the derivative is zero only at $x=0$, the function is strictly increasing. Therefore, the function is one-one.

Finally, we determine if the function is onto by evaluating its range. As $x\to {\frac{\pi }{2}}^{-}$, $\sec x+\tan x\to \infty$, which implies $\log (\sec x+\tan x)\to \infty$, and consequently $f(x)\to \infty$.

As $x\to -{\frac{\pi }{2}}^{+}$, $\sec x+\tan x\to 0^{+}$, which implies $\log (\sec x+\tan x)\to -\infty$, and consequently $f(x)\to -\infty$.

Because the function is continuous and spans from $-\infty$ to $\infty$, its range is $\mathbb{R}$. Thus, the function is onto. We have successfully proven that $f(x)$ is odd, one-one, and onto.