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

1. Nov 22, 2014

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!

2. Nov 22, 2014

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.

Last edited: Nov 22, 2014