# dy/dx=0

• May 7th 2007, 06:17 PM
tummbler
dy/dx=0
hello.. how would i go about solving dy/dx=0?
• May 7th 2007, 06:21 PM
Jhevon
Quote:

Originally Posted by tummbler
hello.. how would i go about solving dy/dx=0?

as in solve for y?

wow, that's a weird question, there's got to be a trick to it that i'm not seeing, i'm going to go ahead and say

dy/dx = 0
=> dy = 0*dx
=> int{1}dy = int{0}
=> y = C where C is a constant
• May 7th 2007, 06:27 PM
ThePerfectHacker
Quote:

Originally Posted by tummbler
hello.. how would i go about solving dy/dx=0?

If y'=0 on I some open interval.

Then, y=C for some real number C.

Proof: Let a<b be in I. Then consider [a,b]. The function y is differenciable fully on [a,b] because it is differenciable on I. Thus, y is continous on [a,b] and as said on (a,b). This tells us that y satisfies Lagrange's Mean Value Theorem. Hence there exists c in (a,b) so that,
f'(c)=(f(b)-f(a))/(b-a)
But f'(c)=0 because f'(x)=0 for all x in I.
Thus,
0=(f(b)-f(a))/(b-a)
Thus,
f(b)=f(a).
This tells us that f is a konstant function on I.
Q.E.D.
• May 7th 2007, 06:28 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
If y'=0 on I some open interval.

Then, y=C for some real number C.

Proof: Let a<b be in I. Then consider [a,b]. The function y is differenciable fully on [a,b] because it is differenciable on I. Thus, y is continous on [a,b] and as said on (a,b). This tells us that y satisfies Lagrange's Mean Value Theorem. Hence there exists c in (a,b) so that,
f'(c)=(f(b)-f(a))/(b-a)
But f'(c)=0 because f'(x)=0 for all x in I.
Thus,
0=(f(b)-f(a))/(b-a)
Thus,
f(b)=f(a).
This tells us that f is a konstant function on I.
Q.E.D.

I guess that's why you're MHF's Number Theorist!
• May 7th 2007, 06:31 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
I guess that's why you're MHF's Number Theorist!

That has nothing to do with number theory. I am far superior in Number/Group/Field Theory than in Analysis. But that proof is a standard. You should know it as well as you know your own name.
• May 7th 2007, 06:34 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
That has nothing to do with number theory. I am far superior in Number/Group/Field Theory than in Analysis. But that proof is a standard. You should know it as well as you know your own name.

well yeah, but i was more refering to our different approaches to the same problem...or did you respond to the problem using a proof because of what i said? (about it being a trick question)
• May 7th 2007, 07:01 PM
tummbler
i should i stated the whole problem. im sorry. it went as follows:

For what value(s) of x in this equation does y attain its minimum value if a, b, and c are all positive constants?

y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

and someone else told me:
Put a=1, b=2, c=3. Then plot the curve. Is the minimum at x=2 or x=0?
-its not.

so, This curve is smooth and y can be minimised by solving dy/dx=0 in the usual manner. I just dont know what the "usual manner" is..