What type of function is f(x) = tan x - x? | vedclass

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(B) Let $f(x) = 	an x - x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(	an x - x) = \

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Always increasing

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(B) Let $f(x) = 	an x - x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(	an x - x) = \sec^2 x - 1$. Using the trigonometric identity $1 + 	an^2 x = \sec^2 x$,we have $\sec^2 x - 1 = 	an^2 x$. Thus,$f'(x) = 	an^2 x$. Since the square of any real number is non-negative,$f'(x) \geq 0$ for all $x$ in the domain. Specifically,$f'(x) > 0$ for all $x 
eq n\pi$ (where $n \in Z$). Since the derivative is non-negative and is not identically zero on any interval,the function is strictly increasing. Therefore,the function is always increasing.

What type of function is $f(x) = \tan x - x$?

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