2^x <= x+1 for x ∈ [0,1]

[emreyunus]

Hi, I tried using mean value theorem but couldn't show 2^2^c < e for c ∈ (0,x). Writing taylor expansion of 2^x also don't work because we need to show 2^x is smaller than something, not greater

 

[Steven G]

You know by now that we do not solve problems for students on this site. Please show us your work as maybe you made a slight mistake and we can get you back on track.

Note that showing 2^2^c < e for c ∈ (0,x) is the same as showing 2^c*ln2 <lne=1 or 2^c < 1/ln2.

Is 2^c increasing?

Actually it seems that you have stated two different problems. Please state one full problems and do not put it into the heading!

 

[emreyunus]

thanks for your help, I am not trying to get my question solved. I am trying to learn it. I am not fully understood that what to do next?

 

[emreyunus]

okay, just see that f''(x)>0, f'(x) and so f(x) are increasing functions. If we write f(1)>f(x) it's done

 

[ashive5i]

Lets say
x = √-1 ......(1)
Doing square on both sides, we get
x^2 = -1

squaring again both sides, we get
x^4 = 1
x = 1
now from eqn 1 above, we can write
1 = √-1
Squaring both sides again - we get
1=-1 ????? How can this be explained in maths ?

 

[emreyunus]

Lets say
x = √-1 ......(1)
Doing square on both sides, we get
x^2 = -1

squaring again both sides, we get
x^4 = 1
x = 1
now from eqn 1 above, we can write
1 = √-1
Squaring both sides again - we get
1=-1 ????? How can this be explained in maths ?

If x^4=1 then lxl=1, its wrong to say x=1

 

[HallsofIvy]

Lets say
x = √-1 ......(1)
Doing square on both sides, we get
x^2 = -1

squaring again both sides, we get
x^4 = 1
x = 1
now from eqn 1 above, we can write
1 = √-1
Squaring both sides again - we get
1=-1 ????? How can this be explained in maths ?

This can be "explained" by pointing out that you made an arithmetic error!

In order to talk about \(\displaystyle \sqrt{-1}\), as you do in the very first line, you must be working in the complex numbers, not the real numbers. In the complex numbers, \(\displaystyle x^4= 1\) does NOT result in \(\displaystyle x= 1\). In the complex numbers, there are 4 possible values for x (even in the real numbers there are 2). x can be 1, -1, i, or -i. You originally defined x to be \(\displaystyle \sqrt{-1}= i\) which IS one of those.