Increasing function

[rtz2k4]

Prove the function is increasing on (0,+inf). Basically I need to prove that the derivate of the function is greater than 0, f'(x)>0, but I get stuck at calculating. I would appreciate a detailed answer.
[math]f:\left(0,\infty \right)\rightarrow R,\:f\left(x\right)=\frac{1}{x+1}-ln\left(x+\frac{3}{2}\right)+ln\left(x+\frac{1}{2}\right)[/math]

 

[Steven G]

If you would appreciate a detailed- or non detailed- answer then you came to the wrong forum.
Please read the forum's posting guideline.
This being a math help forum I am not sure why you would think that you'd get an answer from this site.

 

[Deleted member 4993]

Prove the function is increasing on (0,+inf). Basically I need to prove that the derivate of the function is greater than 0, f'(x)>0, but I get stuck at calculating. I would appreciate a detailed answer.
[math]f:\left(0,\infty \right)\rightarrow R,\:f\left(x\right)=\frac{1}{x+1}-ln\left(x+\frac{3}{2}\right)+ln\left(x+\frac{1}{2}\right)[/math]

Calculating what?

Expression for derivative, or,​

​

The value of the derivative​

 

[rtz2k4]

Calculating what?

Expression for derivative, or,​

​

The value of the derivative​

Expression for derivative

 

[Otis]

I get stuck at calculating.

Hello. Can you show us what you tried? That helps us to get an idea about what you have learned so far.

Do you understand that we differentiate term-by-term?

Have you learned how to differentiate 1/x?

How about differentiating ln(x)?


[imath]\;[/imath]

 

[Deleted member 4993]

Expression for derivative

Do you now that if:

F(x) = A(x) + B(x) +C(x) .................. then .......................

d/dx[F(x)] = d/dx[F(x)] + d/dx[B(x)] + d/dx[C(x)].................. continue......................

 

[rtz2k4]

Do you now that if:

F(x) = A(x) + B(x) +C(x) .................. then .......................

d/dx[F(x)] = d/dx[F(x)] + d/dx[B(x)] + d/dx[C(x)].................. continue......................

Here's my calculations which are wrong, what am I doing wrong?

 

[Deleted member 4993]

View attachment 33918Here's my calculations which are wrong, what am I doing wrong?

How do you know those are wrong ?

 

[Steven G]

Your 2nd to last line contains multiple mistakes. Fix them.
Are you just guessing that last line is less than 0? If not, then show your work.
Your 5th line clearly has a negative sign in front. Where did it come from and why isn't it there in the following lines?

 

[JeffM]

A couple of hints.

You have two natural log functions. Can you reduce them to a single log function before doing differentiation?

If f(x) = g(x) + h(x), then f’(x) = g’(x) + h’(x). If you do differetiation of a complicated function in separate pieces, your chances of copying something wrong go down.

It is not quite correct that f(x) is increasing on (a, b) entails that f’(x) > 0 on (a, b). (It is true that if f’(x) > 0 on (a, b), then f(x) is necessarily increasing on (a, b).) Think about cases where f’(x) = 0, but f(x) is not a local extremum. This is a subtle problem.

 

[Steven G]

A couple of hints.

You have two natural log functions. Can you reduce them to a single log function before doing differentiation?

Jeff,
I disagree with you on this one. Separating logs guarantee us that we never have to use the quotient rule, product rule or general power rule. Not having to use this rules definitely makes finding the derivative easier, if not much easier. If the logs are already separated, then you're lucky.
Steven

 

[BigBeachBanana]

I'll pick it up here.



[math]-\frac{2}{2x+3}-\frac{1}{(x+1)^2}+\frac{2}{2x+1}= \frac{-2(x+1)^2(2x+1)-(2x+1)(2x+3)+2(x+1)^2(2x+3)}{(x+1)^2(2x+1)(2x+3)}\\ =\frac{-2-8x-10x^2-4x^3-3-8x-4x^2+4 x^3 + 14 x^2 + 16 x + 6}{(x+1)^2(2x+1)(2x+3)}\\ =\frac{1}{(x+1)^2(2x+1)(2x+3)}[/math]
It's obvious that [imath](x+1)^2>0, (2x+1)>0, (2x+3)>0[/imath] for [imath]x \in (0,\infty)\implies f'(x)>0[/imath]

 

[Steven G]

Knowing that you want the numerator of the derivative to be a constant, how do you make up such a problem in this form?
I suspect that you can start off with f(x) = 1/(x+a) + bln(cx+d) + eln(fx+g) and work it out. My question is there an easier way?

 

[rtz2k4]

Your 2nd to last line contains multiple mistakes. Fix them.
Are you just guessing that last line is less than 0? If not, then show your work.
Your 5th line clearly has a negative sign in front. Where did it come from and why isn't it there in the following lines?

I've messed up there, I multiplied it with the paranthesis and moved the negative terms to the right and positive terms to the left to calculate easier.