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,
But f'(c)=0 because f'(x)=0 for all x in I.
This tells us that f is a konstant function on I.
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?
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..