Is f(x) Constant Based on Given Derivative and Inequality?

[Jamin2112]

I'm somehow not convinced that that represents f'(x) why not f'(y) since writing x as $$y+Δy$$ allows the Newton's quotient to be f'(y).

i.e. $$lim_ {\delta y->0}\frac {f(y+\triangle y)-f(y)}{y+\triangle y -y}≤ \triangle y$$

I worked out all the kinks. Don't worry.

 

[I like Serena]

Choose y such that 0 < y < ∞. The function f(y)/y is continuous on [0, y] and differentiable on (0, y), so there exists an x in (0, y) such that (f(x)/x)' = [f(y) - f(0)]/[y - 0] = f(y) / y. Am I close?

Close yes.
It should be: f'(x) = [f(y) - f(0)]/[y - 0] = f(y) / y

 

[Jamin2112]

Close yes.
It should be: f'(x) = [f(y) - f(0)]/[y - 0] = f(y) / y

Got it.

Choose x with 0 < x < ∞. The function f is continuous on [0, x] and differential on (0, x). Thus, by the Mean Value Theorem, this exists an x₀ with 0 < x₀ < x and

f(x) = f(x) - f(0) = f '(x₀) * (x - 0) = x * f '(x₀).

Because f ' is monotonically increasing and x > 0, we have x * f '(x₀) ≤ x * f '(x) and accordingly x * f '(x) ≥ f(x), as desired.

http://collegestudybreak.files.wordpress.com/2010/07/success.jpg