DRAW A GRAPH !!!
somehow a smooth curve has to get from 4 at -3 back to 4 at +4
It could be a horizontal line, but that is a trivial case
It could start up, but then would have to come down. SO HAS TO HAVE SLOPE of ZERO in here at least once
if it started down, it has to come up[
so HAS TO GO THROUGH ZERO SLOPE THAT WAY TOO ;)
In fact, I am sure our book tells you at a curve with continuous slope going from A to B has to have slope of
(Yb-Ya)/(Xb-Xa)
at least once between
so
f'(c) = 0
If f(x) is differentiable for the closed interval [-3, 2] such that f(-3) = 4 and f(2) = 4, then there exists a value c, -3 < c < 2 such that
f(c) = 0
f '(c) = 0
f (c) = 5
f '(c) = 5
Not really sure where to start here. No textbook and my teacher never explained it
4 answers
Here you go
http://www.sosmath.com/calculus/diff/der11/der11.html
(mean value theorem)
http://www.sosmath.com/calculus/diff/der11/der11.html
(mean value theorem)
looks like the Mean Value Theorem
Here is a good video by Khan, where he explains your problem is very simple terms
https://www.khanacademy.org/math/differential-calculus/derivative-applications/mean_value_theorem/v/mean-value-theorem
Here is a good video by Khan, where he explains your problem is very simple terms
https://www.khanacademy.org/math/differential-calculus/derivative-applications/mean_value_theorem/v/mean-value-theorem
Maybe you have heard of Rolle's Theorem, a special case of the MVT.