Question

State whether $f(x)=\tan x-x$ is increasing or decreasing in its domain.

Answer 1

Hint: For the given function to say it is increasing or decreasing we have to do the first derivative. After doing the first derivative if the value is > 0 that means the function is increasing or else decreasing.

Complete step-by-step solution -
Given function is $f(x)=\tan x-x$
By doing the first derivative if the value is,
${{f}^{1}}\left( x \right)={{\sec }^{2}}x-1$. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . (1)
We know that \sec x\ge 1$$$$\forall x\in R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
$\forall x\in R$ is the domain.
By squaring on both sides of (2) we get,
{{\sec }^{2}}x\ge 1$$$$\forall x\in R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
By substituting (a) in (1) we get
{{f}^{1}}\left( x \right)={{\sec }^{2}}x-1$$$$\ge 0
From the above we can conclude that
{{f}^{1}}\left( x \right)$$$$\ge 0$$$$\forall x\in R
Therefore $f(x)$ is increasing in the domain.

Note: This is a direct problem which is solved by taking the first derivative and knowing the domain of the function. From (2) we can say that it is a direct property of $\sec x$. If ${{f}^{1}}\left( x \right)$ is greater than zero we say the function is increasing.