I thought I had an answer but was wrong. Anybody?
My best guess would be f(c) = 4, assuming c = 2, but it's just a guess.
Can you imagine the power of an infinite number of Theremins, bringing down defenses, shattering eardrums.
Somehow "piercing" and "woo-woo-woo" don't quite go together in my mind.
The function f is differentiable on the interval 0 < x < 4. If f(1)=1 and f(3)=7, then for some 1 < c < 3, f'(c) must be equal to:
1, 2, 3, 4, or 6?
3. It's the mean value theorem: for some value on the interval, the slope must be equal to the average slope, which is (7-1)/(3-1) = 3.
Jon, that is just about the sweetest story ever. Even with the Tex Avery moments, just so damn sweet. Millions of congratulations all over again!
haha, I loved the Tex Avery moments.
But fate had other plans and now we have two anniversaries to remember.
I tend to get around this problem by just not remembering them.
OK, there are an infininte number of functions that would fulfill that criteria. The function could be a straight line with slope of 2 3. If not, then say it is a function that has a slope of less than 2 3 for part of that interval. Then another part must have a slope of greater than two, and since it's differentiable then there must also be a point with a slope of 2 3 in there....
3. It's the mean value theorem: for some value on the interval, the slope must be equal to the average slope, which is (7-1)/(3-1) = 3.
Argh!
I get the concept right but flunk on the basic math part!!!!
eta: I drew a little graph but made a mistake - I plotted f(1) = 3 for some reason.....