Application of Derivatives...

Problem

Show that the function f(x) = 2x + sin x , x $\in $ R is a strictly increasing function.

Question

Application of Derivatives... 

Problem

Show that the function f(x) = 2x + sin x , x $\in $ R is a strictly increasing function.

mathslover · Answer

@BSwan and @ganeshie8

mathslover · Answer

should I differentiate it?

anonymous · Answer

the definition :D

mathslover · Answer

f'(x) = 2 + cos x ( $\forall x \in R$ )

anonymous · Answer

write it , and go through it

mathslover · Answer

Didn't get you? Did you mean to say to go through the definition?

anonymous · Answer

yep

mathslover · Answer

Well,

this is what I think - 

-1 $\le$ cos x $\le$ 1 

so, 

1 $\le$ 2 + \cos x $\le$ 3

or

1 $\le$ f'(x) $\le$ 3

mathslover · Answer

And for the definition : 

A function f is said to be strictly increasing function in an interval R , iff $x_1 < x_2$ $\implies$ $f(x_1) < f(x_2) $ $\forall x_1 , x_2 \in R $

anonymous · Answer

that tells nothing about  strictly increasing  ,
this tells u that the function is possitive and have upper/lowe bound

anonymous · Answer

dnt forget u have x

mathslover · Answer

And for the TEST of Monotonicity (for Strictly increasing functions : )

f(x) is said to be striclty increasing function in (a,b) iff f'(x) > 0 $\forall$ x $\in$ (a,b) .

anonymous · Answer

ic.. nw , u sued the   differentiate

mathslover · Answer

Well, if I use the Test for Monotonicity , isn't it very clear that , as f'(x) > 0 (for all x belonging to R) ,so, the given function is strictly increasing... ?

anonymous · Answer

your correct !

mathslover · Answer

Oh well... Yeah! Thanks. Probably the easiest one I posted yet...!

anonymous · Answer

yep lol ur the master of Qn XD

mathslover · Answer

More to come... :)

anonymous · Answer

my  Qn sadly no one answer them xD
so feels jelouse

mathslover · Answer

I'm sorry for that.