If the function $f:R ightarrow R$ be defined by $f(x) = an x - x$, then $f(x)$

Question

If the function $f:R ightarrow R$ be defined by $f(x) = 	an x - x$, then $f(x)$

Dr. Abubacker · Accepted Answer

(a)
$\because\ \ f(x) = 	an x - x$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \sec^{2}x - 1 = \frac{1}{\cos^{2}x} - 1 = \frac{1 - \cos^{2}x}{\cos^{2}x}$
Since, $0 \leq \cos^{2}x \leq 1$ for all values of $x$
$	herefore f^{'}(x) > 0$for all values of$x$. Thus, $f(x)$ always increases

Q. If the function $f:R \rightarrow R$ be defined by $f(x) = \tan x - x$, then $f(x)$

Explanation

Reviewed by Dr. Abubacker

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