I am unable to understand what I am asked to do

[George Saliaris]

Find the minimum value of

G(a, b) = (a-2b) ^2 + ( ln((e^a) +1)- b)^2
for the various values of a, b Ε R
A little bit of backround. : I was given some conditions which led to finding a function f(x) = ln(( e^x) +1),I had to prove that f(x) >= 1/2 x +ln2... Anyway.. Can't I just use some transformed inequality of x^2 +y^2 >= 2xy and find the minimum value of G(a,b) in terms of a, b.. Is there something that I am missing?

 

[Dr.Peterson]

We may need to see the whole problem that this is part of; I can't see the connection to the "background".

I would expect to use partial derivatives to find minima for G, just as I'd use derivatives to prove your fact about f (by finding the minimum of f(x)-x/2).

Why do you want to do it a different way? What, exactly, were you asked or told to do (as opposed to what you think you need to do)?

 

[HallsofIvy]

Find the minimum value of

G(a, b) = (a-2b) ^2 + ( ln((e^a) +1)- b)^2
for the various values of a, b Ε R

This doesn't make sense. You say find the minimum values "for the various values f a and b".

There is ONE value of G for any specific values of a and b. You mean, I suspect, find the pairs (a, b) that make G(a, b) minimum.

A little bit of backround. : I was given some conditions which led to finding a function f(x) = ln(( e^x) +1),I had to prove that f(x) >= 1/2 x +ln2... Anyway.. Can't I just use some transformed inequality of x^2 +y^2 >= 2xy and find the minimum value of G(a,b) in terms of a, b.. Is there something that I am missing?

The partial derivatives of G with respect to a and b are
\(\displaystyle G_a= 2(a- 2b)+ 4(ln(e^{2a+ 1)- b)e^{-(2a+ 1)}\) and
\(\displaystyle G_b= -4(a- 2b)\).

Setting those equal to 0 give, from -4(a- 2b)= 0, a= 2b.
With a= 2b, the first equation becomes
\(\displaystyle 4(ln(e^{2a+1}+ 1)- a/2)e^{-(2a+ 1)}= 0\).

Since \(\displaystyle e^{-2a+1}\) is never 0, we must have \(\displaystyle ln(e^{2a+1}+ 1)- a/2= 0\) so \(\displaystyle ln(e^{2a+1}+1)= a/2\) and then \(\displaystyle e^{2a+ 1}+ 1= e^{a/2}\), \(\displaystyle e^{2a+ 1}= e^{a/2}- 1\),

\(\displaystyle e^{2a+ 1}- e^{a/2}= 1\).

 

[George Saliaris]

We may need to see the whole problem that this is part of; I can't see the connection to the "background".

I would expect to use partial derivatives to find minima for G, just as I'd use derivatives to prove your fact about f (by finding the minimum of f(x)-x/2).

Why do you want to do it a different way? What, exactly, were you asked or told to do (as opposed to what you think you need to do)?

Let f : R ->R a differentiable function such that
1) f'(x) = 1 - e^(-f(x)) for every x E R
2) the graph of f(x) is tangent to y = 1/2 x + ln2


a) Prove that f(x) = ln(e^x +1)
b) Show thrat every saight line has at most 2 common points with f(x)
c) Prove that f(x) >= 1/2 x + ln2 x E R
d) (I am translating word by word) Find the minimum value of the expression


G(a,b) = (a-2b) ^2 + ( ln((e^a) +1)- b)^2
for the various values of a, b Ε R

( It's the 4th exercise of the last year's final exam in Calculus 1 in my school)

This doesn't make sense. You say find the minimum values "for the various values f a and b".

There is ONE value of G for any specific values of a and b. You mean, I suspect, find the pairs (a, b) that make G(a, b) minimum.


The partial derivatives of G with respect to a and b are
\(\displaystyle G_a= 2(a- 2b)+ 4(ln(e^{2a+ 1)- b)e^{-(2a+ 1)}\) and
\(\displaystyle G_b= -4(a- 2b)\).

Setting those equal to 0 give, from -4(a- 2b)= 0, a= 2b.
With a= 2b, the first equation becomes
\(\displaystyle 4(ln(e^{2a+1}+ 1)- a/2)e^{-(2a+ 1)}= 0\).

Since \(\displaystyle e^{-2a+1}\) is never 0, we must have \(\displaystyle ln(e^{2a+1}+ 1)- a/2= 0\) so \(\displaystyle ln(e^{2a+1}+1)= a/2\) and then \(\displaystyle e^{2a+ 1}+ 1= e^{a/2}\), \(\displaystyle e^{2a+ 1}= e^{a/2}- 1\),

\(\displaystyle e^{2a+ 1}- e^{a/2}= 1\).

We have not learned anything about partial derivatives,though..