If g is diff'able at x_0, g(x) <= mx + b, show g'(x_0) = m

[rayge]

Pretty much as in the title, except one major condition: g(x_0) = m*x_0 + b.

So, conditions are:
g defined on the reals and is differentiable at x_0.
g(x_0) = m*x_0 + b
mx + b <= g(x) for all x

Then show m = g'(x_0)

I would love to use the intermediate value theorem, or extreme value theorem, or mean value theorem, but since we can only say g is differentiable at x_0 those can't apply since we don't have any intervals where those theorems would apply. Just wondering if there's an obvious theorem to apply here. Thanks!

 

[gopher_p]

In the event that $g'(x_0)=m$, then $y=mx_0+b$ is the equation of the line tangent to the graph of $g$ at the point $(x_0,g(x_0))$.

That's a hint. You should verify that it's true. The hint is not part of a proof, but rather a nudge in the direction of an idea that might lead to a proof.

Edit: There is also a slick way to get this done using Fermat's theorem if you're clever enough to find the right function to use it on.