Maximizing function with derivative constraint - GRE Mathematics Subject Test

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[@devotion7890]

Well explained. I solved it by applying the Mean value theorem: Would f(0) be greater than 8, then the slope of the secant between the two points would be less than -1 and the Mean Value Theorem states that there must be a point in between that has the same slope (i.e. less than -1) and that would be a contradiction to the assumption.

[@mathoutloud]

That’s a great way of doing this! I didn’t think of that!

[@amritlohia8240]

@@mathoutloud The mean value theorem approach is also superior as it only uses the fact that f is differentiable, whereas your approach using the fundamental theorem of calculus has to assume that f has an integrable derivative. That happens to be true here since we're told that f is *continuously* differentiable (and continuous functions are of course integrable), but as the mean value theorem approach shows, that condition actually isn't required.

[@mathoutloud]

@amritlohia8240 the mean value theorem also requires some kind of regularity on the derivative, does it not?

[@amritlohia8240]

@@mathoutloud No, all it requires is differentiability (more precisely, the function is continuous on a closed interval [a,b] and differentiable on the open interval (a,b)) - you can easily look this up in any real analysis book, or even Wikipedia (en.wikipedia.org/wiki/Mean_value_theorem)!

[@mathoutloud]

@amritlohia8240 right, just read that myself. I thought it had to be C^1. Good to know.

[@kiloperson5680]

How did you come up with f(0)

[@mathoutloud]

If you look at the diagram I have around 1:30 you should see why this relation holds. It comes from the bound on the derivative.