Question

How could I prove this? Would this be using a theorem from real analysis?

Answer 1

`"Use the definition of derivative :"`

`f'(x) = lim_{h->0} (f(x+h) - f(x))/h`

`"Here we have"`

`f'(x_0) = lim_{h->0} (f(x_0 + h) - f(x_0))/h`
`g'(x_0) = lim_{h->0} (g(x_0 + h) - g(x_0))/h`

`"We need to prove that"`

`f'(x_0) = g'(x_0)`
`"or"`
`f'(x_0) - g'(x_0) = 0`
`"or"`
`h'(x_0) = 0`
`"with "h(x) = f(x) - g(x)`
`"or"`
`lim_{h->0} (f(x_0 + h) - g(x_0 +h) - f(x_0) + g(x_0))/h = 0`
`"or"`
`lim_{h->0} (f(x_0 + h) - g(x_0 + h))/h = 0`
`"(due to "f(x_0) = g(x_0)")"`

`"Now"`

`f(x_0 + h) <= g(x_0 + h)`
`=> lim <= 0 " if "h>0" and "lim >= 0" if "h < 0`

`"We made the assumption that f and g are differentiable"`
`"so "h(x) = f(x) - g(x)" is also differentiable,"`
`"so the left limit must be equal to the right limit, so"`

`=> lim = 0`
`=> h'(x_0) = 0`
`=> f'(x_0) = g'(x_0)`

Answer 2

I will provide a quicker solution than the one in https://socratic.org/s/aQZyW77G. For this we will have to rely on some familiar results from calculus.

Explanation:

Define `h(x) = f(x)-g(x)`

Since `f(x)\le g(x)`, we have `h(x) le 0`

At `x=x_0` , we have `f(x_0) = g(x_0)`, so that `h(x_0) = 0`

Thus `x=x_0` is a maximum of the differentiable function `h(x)` inside the open interval `(a,b)`. Thus

`h^'(x_0) = 0 implies`

`f^'(x_0)-g^'(x_0) implies`

`f^'(x_0) = g^'(x_0)`