Graphs to prove inequalities (1 Viewer)

[QZP]

Q: Prove tan x > x for 0 < x < pi/2

Am I allowed to just graph the two and show it? My only problem with this method is that it is not really proving it (if I didn't know better, I could have just purposely constructed the two graphs such that tan x > x since I know that's what the question wants)

 

[photastic]

Yea graphing will suffice. Perhaps add in how tanx increases in that domain to further show it's greater than x.

 

[iStudent]

Yea graphing will suffice. Perhaps add in how tanx increases in that domain to further show it's greater than x.

I don't think you can use graphing. Otherwise you can knock down all inequalities questions involving calculus/differentiation by drawing a graph and simply assuming that one lies beneath the other.
It works, but you can't really assume x just happens to lie beneath tanx. I think you need to show it by taking the derivative of both etc.

 

[mreditor16]

Q: Prove tan x > x for 0 < x < pi/2

Am I allowed to just graph the two and show it? My only problem with this method is that it is not really proving it (if I didn't know better, I could have just purposely constructed the two graphs such that tan x > x since I know that's what the question wants)

the point your raise in your second paragraph is most valid. and in my view, markers wouldn't give you the mark, since there is the possibility that you fudged it....

what I would do is say the derivative of y=x is 1. therefore, gradient is constant as 1

then for y = tanx, the derivative is sec^2 (x). now for 0 < x < pi/2 , 0 < cos x < 1

thus, per se, 1/0 > 1 / cos x > 1/1

infinity > sec x > 1

i.e. sec x >1

thus, sec^2 (x) > 1

therefore the gradient of y = tanx is always greater than 1.

thus, with both graphs starting off at (0,0), and tanx having a greater gradient compared to y =x, throughout the provided domain.

thus y = tan x will be always greater than y = x, in the stated domain ....

therefore, tan x > x for 0 < x < pi/2

EDIT - Not sure if this would be accepted by markers - this is my method. But imo, it is definitely more concerning than the quick method you mentioned.

 

[photastic]

The answers used graphs idk what is expected.

 

[faisalabdul16]

the point your raise in your second paragraph is most valid. and in my view, markers wouldn't give you the mark, since there is the possibility that you fudged it....

what I would do is say the derivative of y=x is 1. therefore, gradient is constant as 1

then for y = tanx, the derivative is sec^2 (x). now for 0 < x < pi/2 , 0 < cos x < 1

thus, per se, 1/0 > 1 / cos x > 1/1

infinity > sec x > 1

i.e. sec x >1

thus, sec^2 (x) > 1

therefore the gradient of y = tanx is always greater than 1.

thus, with both graphs starting off at (0,0), and tanx having a greater gradient compared to y =x, throughout the provided domain.

thus y = tan x will be always greater than y = x, in the stated domain ....

therefore, tan x > x for 0 < x < pi/2

EDIT - Not sure if this would be accepted by markers - this is my method. But imo, it is definitely more concerning than the quick method you mentioned.

Pretty much how i would do the question. With questions like these you differentiate and show it's an increasing function or something (i think)

However what i do is differentiate and then use the conditions they give to show blah blah

 

[QZP]

The answers used graphs idk what is expected.

Which answers? Do you have source

 

[Carrotsticks]

Unless it is blatantly obvious graphically (in which case it should also be blatantly obvious algebraically), then you should prove it algebraically. A sketch of something non-obvious like that is most certainly not sufficient.

 

[mirachael]

You could let f(x)=tanx-x and use calculus to prove its greater than 0 for x>0

 

[aDimitri]

You could let f(x)=tanx-x and use calculus to prove its greater than 0 for x>0

this is the method i use for pretty much all inequals like this