Question

Stuck on the rest of attached question:

Answer 1

So x and y are constants? and thus f'(y) is constant. How does this show that f is constant?

Answer 2

Do we just say that since, for all x and y, the derivative of f at y is a constant, so therefore f must be a constant function?

Answer 3

No, wait. If that constant derivative were, say 2, then the function f wouldn't be a constant. grrr

Answer 4

@Polrek , u wrote in last step, f'(y), how ?
assuming x=y+h, right ?
now put x=y+h in left and right side, u will get f'(y)=0, implies f is constant

Answer 5

Hmm.
Look.
\[|[f(x) - f(y)]|/(x-y) \le (x-y)\]
(Assuming x>y. ^^)
Now take
\[\lim_{x \rightarrow y} |f(x) - f(y)|/(x-y) \le x-y\]
\[=> |f'(y)| \le 0\]
However modulus cannot be negative.
Therefore, f'(y) = 0

Answer 6

^^ is a better explanation

Answer 7

Use Intermediate Value theorem
if a function f is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that\[f'(c) =\frac{f(b) - f(a)}{b-a}\]

Answer 8

What your computation actually say that f'(c) < arbitrary small (chosen) number (x+d) - x

Answer 9

Actually it is is simpler just to use the normal Talor linear approximation of order1

Answer 10

Sorry @hartnn, I should've taken the limit as x approaches y in the last step, then said lim(x->y) of (f(x)-f(y))/(x-y), which is the definition of the derivative at y.

So guys, we get \[0\le f'(y)\le0 \], so by the squeeze theorem, f'(y) is 0, and thus f(y) is constant.

Answer 11

yup

Answer 12

Comapring with your requirement it means (by choosing \[ \Delta x \] small enough f'(x) =0

Answer 13

Yep! Thanks guys, you rock!