Key Concepts and Formulas

• Composite Functions: The composition of functions $(h \circ f \circ g)(x)$ is defined as $h(f(g(x)))$. To evaluate this, we apply the functions sequentially from right to left: first $g$ to $x$, then $f$ to the result of $g(x)$, and finally $h$ to the result of $f(g(x))$.
• Trigonometric Identity for Tangent: The identity $\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta}$ is crucial for simplifying expressions involving the form $\frac{1 - \tan\theta}{1 + \tan\theta}$.
• Properties of Tangent Function: Key properties include $\tan(-\theta) = -\tan\theta$ and $\tan(\pi - \theta) = -\tan\theta$. These are used to manipulate angles and signs.

Step-by-Step Solution

We are given the functions $f(x) = \sqrt{x}$, $g(x) = \tan x$, and $h(x) = \frac{1 - x^2}{1 + x^2}$, for $x \in (0, 3/2)$. We need to find $\phi(\pi/3)$, where $\phi(x) = ((h \circ f) \circ g)(x)$.

Step 1: Understand the composite function $\phi(x)$. The function $\phi(x)$ is a composition of three functions: $\phi(x) = h(f(g(x)))$. Why this step? This breaks down the complex composite function into a sequence of simpler function applications, allowing us to evaluate it step-by-step.

Step 2: Evaluate the innermost function $g(x)$ at $x = \pi/3$. We are given $g(x) = \tan x$. We need to find $g(\pi/3)$. $g\left(\frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right)$ Why this step? This is the first step in evaluating the composite function at the given point. We start with the input value and apply the innermost function. $g\left(\frac{\pi}{3}\right) = \sqrt{3}$

Step 3: Evaluate the next function $f$ applied to the result of $g(\pi/3)$. We have $f(x) = \sqrt{x}$. We need to find $f(g(\pi/3))$. $f\left(g\left(\frac{\pi}{3}\right)\right) = f(\sqrt{3})$ Why this step? This is the second step in evaluating the composite function. We take the output of the previous step ($g(\pi/3)$) and use it as the input for the function $f$. $f(\sqrt{3}) = \sqrt{\sqrt{3}} = (3^{1/2})^{1/2} = 3^{1/4}$

Step 4: Evaluate the outermost function $h$ applied to the result of $f(g(\pi/3))$. We have $h(x) = \frac{1 - x^2}{1 + x^2}$. We need to find $h(f(g(\pi/3)))$. $\phi\left(\frac{\pi}{3}\right) = h\left(f\left(g\left(\frac{\pi}{3}\right)\right)\right) = h(3^{1/4})$ Why this step? This is the final step in evaluating the composite function at the given point. We take the output of the previous step ($f(g(\pi/3))$) and use it as the input for the function $h$. Substitute $x = 3^{1/4}$ into $h(x)$: $h(3^{1/4}) = \frac{1 - (3^{1/4})^2}{1 + (3^{1/4})^2}$ Simplify the expression: $h(3^{1/4}) = \frac{1 - 3^{2/4}}{1 + 3^{2/4}} = \frac{1 - 3^{1/2}}{1 + 3^{1/2}} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Step 5: Simplify the expression $\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$ using trigonometric identities. The expression obtained in Step 4 is of the form $\frac{1 - a}{1 + a}$. We can relate this to the tangent subtraction formula. Why this step? The options are given in terms of tangent of certain angles, so we need to convert our numerical result into a trigonometric form. We know that $\tan(\pi/4) = 1$. Let's consider the tangent subtraction formula: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ If we set $A = \pi/4$, we get: $\tan\left(\frac{\pi}{4} - B\right) = \frac{\tan(\pi/4) - \tan B}{1 + \tan(\pi/4) \tan B} = \frac{1 - \tan B}{1 + \tan B}$ We have the expression $\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$. We need to find an angle $B$ such that $\tan B = \sqrt{3}$. We know that $\tan(\pi/3) = \sqrt{3}$. So, we can write: $\frac{1 - \sqrt{3}}{1 + \sqrt{3}} = \frac{1 - \tan(\pi/3)}{1 + \tan(\pi/3)}$ This matches the form $\frac{1 - \tan B}{1 + \tan B}$ with $B = \pi/3$. Therefore, $\frac{1 - \sqrt{3}}{1 + \sqrt{3}} = \tan\left(\frac{\pi}{4} - \frac{\pi}{3}\right)$ Calculate the angle: $\frac{\pi}{4} - \frac{\pi}{3} = \frac{3\pi}{12} - \frac{4\pi}{12} = -\frac{\pi}{12}$ So, $\phi\left(\frac{\pi}{3}\right) = \tan\left(-\frac{\pi}{12}\right)$

Step 6: Use trigonometric properties to match the result with the given options. We have $\phi(\pi/3) = \tan(-\pi/12)$. We need to express this in a form that matches the options. Why this step? The options do not have negative angles. We use the property $\tan(-\theta) = -\tan\theta$ and then relate it to $\tan(\pi - \theta)$. Using $\tan(-\theta) = -\tan\theta$: $\tan\left(-\frac{\pi}{12}\right) = -\tan\left(\frac{\pi}{12}\right)$ Now, we use the property $\tan(\pi - \theta) = -\tan\theta$. Let $\theta = \pi/12$. $\tan\left(\pi - \frac{\pi}{12}\right) = -\tan\left(\frac{\pi}{12}\right)$ $\tan\left(\frac{12\pi - \pi}{12}\right) = -\tan\left(\frac{\pi}{12}\right)$ $\tan\left(\frac{11\pi}{12}\right) = -\tan\left(\frac{\pi}{12}\right)$ Therefore, $\phi\left(\frac{\pi}{3}\right) = \tan\left(\frac{11\pi}{12}\right)$

Step 7: Identify the correct option. Comparing our result with the given options: (A) $\tan \frac{7\pi}{12}$ (B) $\tan \frac{11\pi}{12}$ (C) $\tan \frac{\pi}{12}$ (D) $\tan \frac{5\pi}{12}$ Our calculated value $\tan\left(\frac{11\pi}{12}\right)$ matches option (B).

Common Mistakes & Tips

• Order of Composition: Ensure the order of function application is correct: $h(f(g(x)))$ means $g$ is applied first, then $f$, then $h$.
• Domain and Range: While not strictly necessary for this specific evaluation, always be mindful of the domains and ranges of composite functions. For instance, if $g(x)$ yielded a negative value, $f(x) = \sqrt{g(x)}$ would be undefined. In this problem, $x \in (0, 3/2)$ which is within $(0, \pi/2)$, so $\tan x > 0$.
• Trigonometric Identity Application: Recognize the form $\frac{1 - \tan\theta}{1 + \tan\theta}$ and its relation to $\tan(\pi/4 - \theta)$.

Summary

The problem requires evaluating a composite function $\phi(x) = ((h \circ f) \circ g)(x)$ at a specific point $x = \pi/3$. We first computed $g(\pi/3)$, then applied $f$ to this result, and finally applied $h$ to obtain a numerical value. This numerical value was then simplified using trigonometric identities to match one of the given options. The key was recognizing the form $\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$ and relating it to the tangent subtraction formula.

The final answer is $\boxed{\tan \frac{11\pi}{12}}$.