To determine if the function
f(x)=2x−tan−1(x)−log(x+√x2+1)is monotonically increasing, we need to evaluate the derivative of the function and find the range of
x for which the derivative is non-negative (i.e.,
f′(x)≥0 ).
Let's compute the derivative
f′(x) :
f′(x)=[2x−tan−1(x)−log(x+√x2+1)]Using the chain rule and known derivatives:
The derivative of
2x is 2 .
The derivative of
tan−1(x) is
.
For the derivative of
log(x+√x2+1) :
Let
u=x+√x2+1. Then
=1+Therefore,
Combining these, the derivative
f′(x) becomes:
f′(x)=2−−Next, let's simplify
:
This fraction simplifies to
=1, since numerator and denominator become equal with algebraic manipulation.
Thus,
f′(x) becomes:
f′(x)=2−−1=1−Let's evaluate for which values of
x this derivative is non-negative:
This inequality always holds for all real numbers. Thus,
f′(x)≥0 for all x∈ℝTherefore, the function
f(x) is monotonically increasing for all
x in
ℝ. Hence, the correct option is:
Option C
x∈ℝ