Inequality (1 Viewer)

[pomsky]

Yo guys!
How do you work with differentiation in proving stuff in inequalities? I just don't get it...

Q1: Given f(x)= ln(x)/x
a.) Show that e^pi > pi^e

Can you just sub stuff in?

Q2: a.) Show that sin x < x for x> 0.
NB: Please assume I don't know any official theorems. My question in this one is the domain shifts. Why do we take 0>cosx >0 and then assume the equality holds when 1- cosx = 0?

b.) Let g(x) = sinx-x + (x^3)/6. Show that the graph of y= g(x) is concave up for x >0.

c.) By considering the first two derivatives of g(x). Show that sinx > x-(x^3)/6 for x>0

Thanks!

 

[pomsky]

bumpity bump~

 

[pomsky]

bummp

 

[Carrotsticks]

1) What you'll find is that the stationary point of ln(x)/x occurs at (e,1/e), which is a maximum SP (and is also the global max). This means that no matter what x value we substitute in, the output is ALWAYS less than 1/e. So if I substitute x=pi, then I get ln(pi)/pi, which MUST be less than 1/e as 1/e is the maximal output. So ln(pi)/pi < 1/e. Try to manipulate it from there.

2) Think about this... if a function starts off at say (0,0) and we somehow know for sure that it is always increasing for x>0.... then doesn't that mean that the function must then always be ABOVE the x axis for all x>0? Try drawing a quick diagram with the above features to see for yourself.

Now, we don't know how to prove directly that sin(x)<x. However, we DO know how to prove that Consider f(x)=x-sin(x) is always above the x axis. In other words, x-sin(x)>0 and thus the required result. Consider what I said in the above paragraph. Observe that f(0)=0, so it starts off at the origin. Now examine the derivative f'(x)=1-cos(x). Notice that no matter what x is, 1-cos(x) is always positive? This is because the value of cos(x) is NEVER greater than 1, so 1-cos(x) is either positive or equal to zero. But 1-cos(x) is f'(x), so f'(x) is either positive or equal to zero. In any case, we have just shown that f(0)=0 and f'(x) > 0 for all x>0, so we've just proven that f(x)>0 for all x>0. But remember we've defined f(x) to be x-sin(x), so x-sin(x)>0 for all x>0 and therefore x>sin(x) for all x>0.

3) To show that g(x) is concave up for x>0, we need to show that g''(x)>0 for x>0. But observe that g'(x)=cos(x)-1+x^2/2 and so g''(x)=-sin(x)+x=x-sin(x). But remember from earlier when we proved that x-sin(x)>0 for all x>0? We can use that result again to show that g''(x)>0 for all x>0 and thus g(x) is concave up for all x>0.

4) Notice the above inequality is the same as g(x)>0? So if we somehow prove that g(x)>0, then we are pretty much done. Now observe that g(0)=0 so it starts off at the origin. From earlier, we proved that g(x) is always concave up. It's tempting to say "So therefore since g(0)=0 and g(x) is concave up, then g(x) must always be positive then", but that is incorrect (try to find out why). I'll leave it to you to find out what other piece of information we need to guarantee that g(x) is always positive. Read the original question carefully.

 

[Carrotsticks]

Sorry, wall of text. This is something best explained in person versus online.