MVT to prove inequalities (1 Viewer)

[mrpotatoed]

How do you use the MVT to prove inequalities where its x>a>0 instead of just x>0? Like when its x>0 you can just let b=x and a=0 in the MVT formula thingy but not sure what to do for this. Examples are Q 97b and 98a,c,d of the math 1151 calc booklet at unsw. Any help appreciated.

 

[InteGrand]

How do you use the MVT to prove inequalities where its x>a>0 instead of just x>0? Like when its x>0 you can just let b=x and a=0 in the MVT formula thingy but not sure what to do for this. Examples are Q 97b and 98a,c,d of the math 1151 calc booklet at unsw. Any help appreciated.

Not sure what the Q. is. Can you provide an example?

 

[mrpotatoed]

I don't know how to use latex but I'll try to just type it out manually.

Use the MVT to show that:

(x-a)/a>ln(x/a)>(x-a)/x if x>a>0

(sorry all the inequalities are the wrong way around, for someone reason my text disappears if I write it normally)

 

[InteGrand]

I don't know how to use latex but I'll try to just type it out manually.

Use the MVT to show that:

(x-a)/a>ln(x/a)>(x-a)/x if x>a>0

(sorry all the inequalities are the wrong way around, for someone reason my text disappears if I write it normally)

Here's how to write the inequalities normally and avoid the text disappearing:

(x-a)/x < ln(x/a) < (x-a)/a.

I.e. put spaces between the inequality signs and nearby text.

 

[InteGrand]

 

[InteGrand]

 

[leehuan]

I'm just doing the entire question as I am using this as an exercise for myself.

(Essentially, using what InteGrand said that I chose to not read...)
Let f: [a,x]->R, f(t) = ln(t) (where obviously x>a).
As f is continuous on it's domain, and differentiable over (a,x), there must exist at least one c in (a,x)
=> ( f(x) - f(a) )/(x - a) = f'(c)
<=> ( ln(x) - ln(a) )/(x-a) = 1/c

But since a < c < x, we have 1/x < 1/c < 1/a

Hence, 1/x < ln(x/a)/(x - a) < 1/a using a log law
and recalling that x>a
(x-a)/x < ln(x/a) < (x-a)/a

QED

 

[mrpotatoed]

Thanks for the help guys