why does the MVT allow us to conclude that if D(f(x)) = D(g(x)) then f(x) = g(x) + C ???
MIT 18.01 Single Variable Calculus (OCW)
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im assuming D is differential
hey i dont htink thats the way you want it proved tho, with the mean value theorm
go here and watch from time 28:04 mins
I was also wondering the same thing when going through the course. I tried to make up the proof myself, I don't know if it is correct but here it is:
MVT says (f(b)-f(a))/b-a = D(f(c)) and (g(b)-g(a))/b-a = D(g(c))
But if we take 'b' as variable(this is my major assumption) over the domain of the functions f and g, we get
(f(x)-f(a))/x-a = D(f(x)) and (g(x)-g(a))/x-a = D(g(x))
And we are given D(f(x)) = D(g(x)), so
(f(x)-f(a))/x-a = (g(x)-g(a))/x-a
=> f(x)-f(a) = g(x)-g(a)
=> f(x) = g(x) + (f(a)-g(a))
=> f(x) = g(x) + constant; as ((f(a)-g(a)) is a constant
You can prove it without MVT:
Proof by Contraposition:
To prove that p implies q, you can instead attempt to prove NOT q implies NOT p, this is because the two statements are logically equivalent. Thus, if the "NOT" statement is true, then the other statement must be true as well.
In this case, p is D(f(x))=D(g(x)) and q is f(x)=g(x)+c
This means that NOT p is D(f(x)) =/= D(g(x)) and NOT q is f(x)=/=g(x)+c
So lets try to prove NOT q implies NOT p:
if f(x) =/= g(x) + C, then by the definition of the derivative, D(f(x)) =/= D(g(x)).
Therefore, we have proven our contrapositive, and logically, proven the original.