MAT137 Calculus! Lecture 10 Today we will study the Mean Value Theorem and its applications. Extrema. Optimization Problems. (4.2-4.5) PS4 is due this Friday June 23. Next class: Curve Sketching (4.6-4.8)
Mean value theorem (MVT) MVT Let f be a function that satisfies the following hypothesis: 1 f is continuous on the closed interval [a,b] 2 f is differentiable on the open interval (a,b) then there is at least a point c (a,b) such that f (c) = f(b) f(a) b a
Mean value theorem (MVT) remarks MVT Let f be a function that satisfies the following hypothesis: 1 f is continuous on the closed interval [a,b] 2 f is differentiable on the open interval (a,b) then there is at least a point c (a,b) such that f (c) = f(b) f(a) b a Note that the theorem does not tell us how to find the number c. It only tells us that such a number exists.
Mean value theorem (MVT) remarks MVT Let f be a function that satisfies the following hypothesis: 1 f is continuous on the closed interval [a,b] 2 f is differentiable on the open interval (a,b) then there is at least a point c (a,b) such that f (c) = f(b) f(a) b a Geometrical interpretation: see the board
Mean value theorem (MVT) remarks MVT Let f be a function that satisfies the following hypothesis: 1 f is continuous on the closed interval [a,b] 2 f is differentiable on the open interval (a,b) then there is at least a point c (a,b) such that f (c) = f(b) f(a) b a Physical interpretation: if I run continuously for an hour in High Park and I described my trajectory using a function s : [0,1] R, the average velocity will coincide with the instantaneous velocity at some point of my trajectory, i.e., there is a time t 0 [0,1] such that s (t 0 ) = s(1) s(0) 1 0 = s(1) s(0).
Mean value theorem (MVT) remarks MVT Let f be a function that satisfies the following hypothesis: 1 f is continuous on the closed interval [a,b] 2 f is differentiable on the open interval (a,b) then there is at least a point c (a,b) such that f (c) = f(b) f(a) b a It is left as an exercise to find examples of a function f for which the MVT fails when: f is not differentiable on (a,b) but continuous on [a,b] f is differentiable on (a,b) but f is not continuous on [a,b]
Mean value theorem (MVT) Rolle s theorem We will prove the mean value theorem in steps. Rolle s theorem Suppose that f is continuous on the closed interval [a,b] and it is differentiable on the open interval (a,b). If f(a) = f(b) = 0, then there is at least a number c (a,b) for which f (c) = 0. Proof: we will prove this theorem later
Mean value theorem (MVT) Rolle s theorem: remarks Some times Rolle s theorem is stated as Rolle s theorem Suppose that g is continuous on the closed interval [a,b] and it is differentiable on the open interval (a,b). If g(a) = g(b), then there is at least a number c (a,b) for which g (c) = 0. Proof: considering the auxiliary function f(x) = g(x) g(a) and using our version of Rolle s theorem, you can prove this at home! Rolle s theorem is a special case of the mean value theorem (why?).
Mean value theorem (MVT) proof Proof of the Mean Value Theorem: Consider the auxiliary function ( ) f(b) f(a) g(x) = f(x) (x a)+f(a). b a And let s verify that g satisfies the hypothesis of Rolle s theorem (see the board).
Applications of MVT Corollary If f (x) = 0 for all x (a,b), then f is constant on (a,b). Proof: on the board. Desideratum: w (a,b), z (a,b),f(w) = f(z).
Applications of MVT Corollary If f (x) = g (x) for all x (a,b), then there is a constant C R such that f(x) = g(x)+c, for all x (a,b). Proof: exercise.
Applications of MVT Increasing functions Which of the following are valid ways to to write the definition of increasing function? Definition Let f be a function defined on D. We say that f is increasing if... 1 x 1,x 2 D s.t. x 1 < x 2 f(x 1 ) < f(x 2 ). 2 x 1,x 2 D, f(x 1 ) < f(x 2 ). 3 x 1,x 2 D, x 1 < x 2 f(x 1 ) < f(x 2 ). 4 x 1 D, f (x 1 ) > 0.
Applications of MVT Increasing functions Which of the following are valid ways to to write the definition of increasing function? Definition Let f be a function defined on D. We say that f is increasing if... 1 x 1,x 2 D s.t. x 1 < x 2 f(x 1 ) < f(x 2 ). 2 x 1,x 2 D, f(x 1 ) < f(x 2 ). 3 x 1,x 2 D, x 1 < x 2 f(x 1 ) < f(x 2 ). 4 x 1 D, f (x 1 ) > 0.
Applications of MVT Increasing functions Definition Let f be a function defined on an interval I. We say that f is increasing on I if for all x 1,x 2 I, we have x 1 < x 2 f(x 1 ) < f(x 2 ). We say that f is non-decreasing on I if for all x 1,x 2 I, we have x 1 < x 2 f(x 1 ) f(x 2 ). In a similar way, we can define when a function f is decreasing and non-increasing on an interval I (homework).
Applications of MVT Increasing functions vs. derivatives Theorem Let a < b and let f be a function differentiable on (a,b). If f (x) > 0, for all x (a,b), then f is increasing on (a,b). If f (x) < 0, for all x (a,b), then f is decreasing on (a,b). Proof: we will prove only the first statement. We need to prove that x 1,x 2 (a,b),x 1 < x 2 f(x 1 ) f(x 2 ). So, fix x 1,x 2 (a,b). Now we need to prove that: if x 1 < x 2, then f(x 1 ) f(x 2 ) Assume that x 1 < x 2. See the board for the rest of the proof...
Applications of MVT Increasing functions vs. derivatives Example 0 Example 0 Let f(x) = x 3 27x 20. Find the intervals on which f is increasing or decreasing. Find the local extreme values.
Extrema Recall Extreme value theorem. If f is is continuous on a closed bounded interval [a,b], then f attains both a maximum value M and a minimum value m in [a,b]. That is, there are numbers x 1,x 2 [a,b] such that f(x 1 ) = M, f(x 2 ) = M and for all x [a,b], m f(x) M. How can we find M and m?
Extrema Definition Let f be a function with domain D and let c D. We say that f has a maximum (global maximum) at c if f(x) f(c), for all x D. Definition Let f be a function with domain D and let c D. We say that f has a minimum (global minimum) at c if f(c) f(x), for all x D. Maximum and minimum are called extreme values of the function f.
Extrema Example 1 Does f have a maximum if f(x) = x 2? Does it have a minimum? Minimum of f is attain at 0 and it is m = f(0) = 0. There is no maximum.
Extrema Example 2 Does the following function have a maximum or a minimum? x if x < 0 f(x) = sin(x) if 0 x 2π x 2π if x > 2π f has no maximum and it has no minimum.
Extrema Example 2 Does the following function have a maximum or a minimum? x if x < 0 f(x) = sin(x) if 0 x 2π x 2π if x > 2π What about g given by g(x) = sin(x), for x [0,2π]? Note that g = f [0,2π]. g has a maximum at π 2 and a minimum at 3π 2. Even though f has neither a maximum nor a minimum, f does have local max and local min
Local extrema Definition Let f be a function with domain D and let c D. We say that f has a local maximum at c if there is δ > 0 such that x c < δ f(x) f(c). Definition Let f be a function with domain D and let c D. We say that f has a local minimum at c if there is δ > 0 such that x c < δ f(c) f(x).
Local extrema Example 3 Does the following function have a local maximum or a local minimum? x if x < 0 f(x) = sin(x) if 0 x 2π x if x > 2π Solution: see the board.
Local extrema Definition Let f be a function with domain D and let c D. We say that c is an interior point of D if there is an open interval I such that c I and I D. Equivalently, c is an interior point of D if there is δ > 0 such that (c δ,c +δ) D. Theorem If f has a local maximum or local minimum at an interior point c of its domain, then f (c) = 0 or f (c) does not exist.
Local extrema Theorem If f has a local maximum or local minimum at an interior point c of its domain, then f (c) = 0 or f (c) does not exist. Definition The interior points c of the domain of f for which are called critical points. f (c) = 0 or f (c) does not exist.
Finding local and global extrema Let f be continuous on a closed bounded interval [a,b]. The Extreme value theorem says that f attains a maximum and a minimum. The Local Extreme Value Theorem says that the only places where a function can possibly have an extreme are 1 interior points c where f (c) = 0, 2 interior points c where f (c) is undefined, 3 the end points of the domain of f (i.e. a and b).
Finding local and global extrema Example 3 Example 3 Find the local and global extrema of the function f given by f(x) = x 2 3 (x 1) 3 on the interval [ 1,2]. See the board for the solution.
Mean value theorem (MVT) Rolle s theorem Theorem Suppose that f is differentiable at x 0. If f (x 0 ) > 0, then for all positive h sufficiently small. If f (x 0 ) < 0, then for all positive h sufficiently small. Proof: see the chalkboard f(x 0 h) < f(x 0 ) < f(x 0 +h) f(x 0 h) < f(x 0 ) < f(x 0 +h)
Mean value theorem (MVT) Rolle s theorem Rolle s theorem Suppose that f is continuous on the closed interval [a,b] and it is differentiable on the open interval (a,b). If f(a) = f(b) = 0, then there is at least a number c (a,b) for which Proof: see the chalkboard Theorem 2.6 of the book: Extreme value theorem. f (c) = 0. If f is is continuous on a closed bounded interval [a,b], then f attains both a maximum value M and a minimum value m in [a,b]. That is, there are numbers x 1,x 2 [a,b] such that f(x 1 ) = M, f(x 2 ) = M and for all x [a,b], m f(x) M.