# find f(x)=2x + arccotx + log{ [1+x^2]^(1/2) -x} is increasing or decreasingNeed urgent answer

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### 2 Answers

If you have a positive number and you take the log of it, you don't necessarily have a positive number.

E.g.: ln(1/e)=-1

1/e is positive, but -1 is negative.

So if you show [1+x^2]^(1/2) -x>0 you have shown that

log{ [1+x^2]^(1/2) -x} is defined (since you can only take logs of positive numbers) but you haven't shown that number is positive.

Moreover, showing something is positive doesn't mean it is increasing.

`f(x)=2x + "arccot" (x) + "log" (sqrt(1+x^2)-x)`

To find out if this is increasing or decreasing, we take a derivative and see if it is positive or negative:

`f'(x)=2 + (-1)/(1+x^2)+ ((1)/(sqrt(1+x^2)-x))(((1)/(2))(1+x^2)^(-1/2)*2x-1)`

`=2 - (1)/(1+x^2)+(x(1+x^2)^(-1/2)-1)/(sqrt(1+x^2)-x)`

We start by doing some algebra on that last term:

`(x(1+x^2)^(-1/2)-1)/(sqrt(1+x^2)-x)`

`=((x(1+x^2)^(-1/2)-1)(sqrt(1+x^2)+x))/((sqrt(1+x^2)-x)(sqrt(1+x^2)+x))`

`=(x+x^2(1+x^2)^(-1/2)-(1+x^2)^(1/2)-x)/(1+x^2-x^2)`

Putting this back into `f'` we have:

`f'(x)=2-(1)/(1+x^2)+x+x^2(1+x^2)^(-1/2)-(1+x^2)^(1/2)-x`

`=2-(1)/(1+x^2)+(x^2)/(sqrt(1+x^2))-sqrt(1+x^2)`

`=2-(1)/(1+x^2)+(x^2sqrt(1+x^2))/(1+x^2)-(sqrt(1+x^2)(1+x^2))/(1+x^2)`

`=(2(1+x^2)-1+x^2 sqrt(1+x^2)-(1+x^2)sqrt(1+x^2))/(1+x^2)`

`=(1+2x^2-sqrt(1+x^2))/(1+x^2)`

The denominator is always positive, so it won't affect whether `f'` is positive or negative. So consider the numerator:

`1+2x^2-sqrt(1+x^2)=(1+x^2)-sqrt(1+x^2)+x^2`

Now, if `r>1` then `r> sqrt(r)`

`x^2>= 0`, and, unless `x=0` , we have `x^2>0` so `x^2+1> 1`, so `x^2+1> sqrt(x^2+1)`

(and at `x=0` we have `x^2+1=sqrt(x^2+1)` )

thus `(1+x^2)-sqrt(1+x^2)> 0` so that is positive

also, `x^2>0`, so that is also positive.

A positive plus a positive is positive

So, unless `x=0` , the numerator is positive, so `f'(x)` is positive, so `f(x)` is increasing

And at `x=0` the numerator is exactly 0, so `f'(x)=0` , so `f(x)` "flattens out" ` `

Let y=arc cotx

Then coty=x

By differntiation

d(coty)/dx = 1

-sec^2x/tan^2x *dy/dx= 1

-(1+tan^2x)/tan^2x*dy/dx =1

-(cot^2x+1)*dy/dx =1

dy/dx= -1/(1+x^2)

Let P=2x+arc cotx

dP/dx = 2-1/(1+x^2)

= (2x^2+1)/(1+x^2)

For all x ;2x^2+1>0

For all x;1+x^2 >0

So dP/dx >0 for all x.

**dP/dx means the gradient of the function P. Since dP/dx>0 always P is increasing.**

Let Q=log{ [1+x^2]^(1/2) -x}

We know 1+x^2>x^2

Then [1+x^2]^(1/2) >sqrt(x^2)

[1+x^2]^(1/2) >x

Then [1+x^2]^(1/2) -x>0

So log{ [1+x^2]^(1/2) -x}>0 since log is not negative.

So every value of x it is a poitive value and increasing.

So Q is increasing for all x.

But f(x) = P+Q

**Since both P and Q are increasing f(x) is increasing.**

**Sources:**