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
To solve this problem, we can make use of the Intermediate Value Theorem (IVT) and the Mean Value Theorem (MVT), which are fundamental concepts in calculus.
First, let's start with the Intermediate Value Theorem. The IVT states that if a function is continuous on a closed interval [a, b], and it takes on two different values, f(a) and f(b), then it must also take on every value between f(a) and f(b). In other words, if the function is continuous and f(a) < k < f(b), then there exists some value c in the interval (a, b) such that .
In our problem, we know that f(-3) = 4 and f(2) = 4. We want to find a value c such that f(c) = 0. Since f(-3) = 4 and f(2) = 4 (both equal values), we can use the IVT to conclude that there exists some value c in the interval (-3, 2) such that f(c) = 0. Therefore, the statement "f(c) = 0" is true.
Now let's move on to the Mean Value Theorem. The MVT states that if a function is differentiable on an open interval (a, b), then there exists at least one value c in that interval such that f'(c) = (f(b) - f(a))/(b - a). In simpler terms, it guarantees that there exists some value c in the interval (a, b) where the derivative of the function equals the slope of the secant line between the points (a, f(a)) and (b, f(b)).
In our problem, we are given that f(x) is differentiable for the closed interval [-3, 2]. Since the interval is closed, we will use the MVT for closed intervals, which states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists some value c in the interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
We want to find a value c such that f'(c) = 0. By applying the MVT to the interval (-3, 2) and using the fact that f(-3) = 4 and f(2) = 4 (both equal values), we can conclude that there exists some value c in the interval (-3, 2) such that f'(c) = 0. Therefore, the statement "f'(c) = 0" is true.
For the statements "f(c) = 5" and "f'(c) = 5", there is no guarantee that such values exist based on the given information of f(-3) = 4 and f(2) = 4. We cannot make those conclusions because the IVT and MVT don't provide evidence for those cases.
In summary, the statements "f(c) = 0" and "f'(c) = 0" are true given the information provided. The statements "f(c) = 5" and "f'(c) = 5" cannot be determined solely based on the given information.
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
Maybe you have heard of Rolle's Theorem, a special case of the MVT.
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
Here you go
http://www.sosmath.com/calculus/diff/der11/der11.html
(mean value theorem)