Step 1: Concept Clarification:
To ascertain if a function is increasing or decreasing over a specified interval, one must examine the sign of its first derivative on that interval. A positive first derivative (f'(x)>0) indicates an increasing function, while a negative first derivative (f'(x)<0) signifies a decreasing function.
Step 2: Methodology:
1. Calculate the derivative of the function f(x), denoted as f'(x).
2. Evaluate the sign of f'(x) within the interval [0, $\frac{\pi}{2}$).
Step 3: Derivation and Analysis:
The function provided is f(x) = tanx - x.
The first derivative, f'(x), is calculated as follows:
\[ f'(x) = \frac{d}{dx}(\tan x - x) = \sec^2 x - 1 \]
Utilizing the trigonometric identity $\sec^2 x = 1 + \tan^2 x$, we can simplify f'(x):
\[ f'(x) = (1 + \tan^2 x) - 1 = \tan^2 x \]
Next, we analyze the sign of f'(x) = $\tan^2 x$ on the interval [0, $\frac{\pi}{2}$).
Within this interval, tanx is defined for all x. Since $\tan^2 x$ represents a squared value, it is inherently non-negative.
\[ f'(x) = \tan^2 x \ge 0 \]
The derivative f'(x) equals zero only at x = 0, as tan(0) = 0. For all other values of x in the interval (0, $\frac{\pi}{2}$), tanx is positive, resulting in $\tan^2 x$ being strictly positive.
Given that f'(x) $\ge$ 0 on the interval [0, $\frac{\pi}{2}$) and is not zero across any subinterval, the function f(x) demonstrates increasing behavior on this interval.
Step 4: Conclusion:
The function f(x) = tanx - x is classified as an increasing function on [0, $\frac{\pi}{2}$).