Question

A problem based on mean value theorem. Please ping me.
1) Suppose that f'(x) >= 0 for all x > a. Prove that f(x) >= f(a) for all x >= a?

Answer 1

if x = a, we have satisfied the requirement (obviously)

if x > a then by the MVT \(\exists\) c, \(a<c<x\) s.t.\[f'\left( c \right)=\frac{ f \left( x \right)-f \left( a \right) }{ x-a }>0\]


let me knwo if you need the rest

Answer 2

assuming f is continuous on the interval [a, x]

Answer 3

Can u send me the full problem in detail please

Answer 4

@pgpilot326

Answer 5

since f'(x) \(\ge\) 0, \(\forall\) x > a
=> f(x) - f(a) \(\ge\) 0 (because x - a > 0)
=> f(x) \(\ge\) f(a)