Shape of a curve. Nov 15, 2016
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1 Shape of a curve Nov 15, 2016
2 y = f(x) Where does the curve of f attain its maximum or minimum value? Where does the curve of f increase or decrease? What is its sketch?
3 Some definitions Def: Absolute maximum f(c) if f(c) f(x) for all x in domain of f Def: Absolute minimum f(c) if f(c) f(x) for all x in domain of f
4 Some definitions Def: Absolute maximum f(c) if f(c) f(x) for all x in domain of f Def: Absolute minimum f(c) if f(c) f(x) for all x in domain of f Note: abs max/min can occur at multiple points, e.g. sin(x). Local maximum/minimum: similar definition restricted to an open interval containing c.
5 Do all functions have extreme values?
6 Do all functions have extreme values? No! E.g. y = x, y = x 3, y = e x, y = 1/x no abs max/min y = x 2, y = x abs min, no max y = x 2 on [-1, 1] has absolute max y(-1)=y(1)=1.
7 Extreme Value Theorem Thrm: If f is continuous on [a,b], then f attains max/min values.
8 Extreme Value Theorem Thrm: If f is continuous on [a,b], then f attains max/min values. What if f is not continuous? What if the interval is open?
9 Properties of extreme values Thrm: If f has a local max or min at x=c and f (c) exists, then f (c) = 0. (Fermat s theorem)
10 Properties of extreme values Thrm: If f has a local max or min at x=c and f (c) exists, then f (c) = 0. (Fermat s theorem) Does this mean that if f (a)=0, then f(a) is an extreme value?
11 How can we infer the shape of a graph? We know f (a) is slope of curve at a point x=a. Suppose we know f (a) >0, positive slope, increasing f. Can we claim f (x) >0 for all x close to a, i.e. can we claim f is increasing everywhere near a?
12 How can we infer the shape of a graph? We know f (a) is slope of curve at a point x=a. Suppose we know f (a) >0, positive slope, increasing f. Can we claim f (x) >0 for all x close to a, i.e. can we claim f is increasing everywhere near a? Key question: How can we connect the local picture (slope at a point) to global picture (average slope and shape of graph across an interval).
13 Mean Value Theorem Thrm: If f is continuous on [a, b], and if f is differentiable on (a, b), then there is a number c in (a, b) such that f(b) f(a) = f (c) b a Average slope across an interval = slope at some point inside the interval
14 Problem: Find all numbers c that satisfy the conclusion of the MVT for y = x 3-3x + 1 on [-2, 2].
15 Problem: Find all numbers c that satisfy the conclusion of the MVT for y = x 3-3x + 1 on [-2, 2]. Solution: f(x)= x 3-3x + 1 is differentiable everywhere, therefore continuous. By MVT there is a number c in (-2, 2) such that [f(2)-f(-2)]/(2+2) = f (c) = 3c 2 3 [8-6+1 (-8+6+1)] /4 = 3c 2 3 1= 3c 2 3 c 1 = 2/3 1/2 and c 2 = - 2/3 1/2
16 The truth is nobody cares about the exact value of c. Its existence is what matters!
17 Does the MVT make sense intuitively? It takes you two hours to drive from Vancouver to Hope (about 150 km east of Vancouver). Your average speed for the trip is 150/2 = 75 km/hr. Is it reasonable to claim that your speed was exactly 75 km/hr at some point during the trip? Why?
18 Careful at applying the theorem y = x-1 on [-1,4] though [f(4)-f(-1)]/(4+1) = 1/5 = f (c)??? The function is not differentiable at x=1.
19 Mean Value Theorem Thrm: If f is continuous on [a, b], and if f is differentiable on (a, b), then there is a number c in (a, b) such that f(b) f(a) = f (c) b a What if f is not diff at end points? What if f is not diff at an interior point? What if f is not continuous at endpoints? What if f is not continuous at an interior point?
20 Proof of MVT: Preliminary work (Rolle s) Theorem: (special case of MVT) If f is continuous on [a, b], If f is differentiable on (a, b), If f (a) = f (b), then there is a number c in (a, b) where f (c) = 0.
21 Rolle s theorem: proof I) If f(x) = k (constant function), then f =0 everywhere
22 Rolle s theorem: proof I) If f(x) = k (constant function), then f =0 everywhere II) If f(x) is not a constant, then it must reach its maximum (or minimum) at some point c in (a,b) by Extreme Value Theorem (f is continuous on closed interval). Because f(c) is a maximum (or minimum), f (c) =0 by Fermat s theorem (at an extremum the slope is zero).
23 MVT: proof The idea is to tilt the graph back to Rolle s special case.
24 MVT: proof The idea is to tilt the graph back to Rolle s special case. Equation of secant line: y = f(a) + (Δf/Δx)(x-a)
25 MVT: proof The idea is to tilt the graph back to Rolle s special case. Equation of secant line: y = f(a) + (Δf/Δx)(x-a) Vertical distance between f and secant line is G(x)= f(x) - f(a) - (Δf/Δx)(x-a) G(x) is continuous on [a, b], differentiable on (a, b),
26 MVT: proof The idea is to tilt the graph back to Rolle s special case. Equation of secant line: y = f(a) + (Δf/Δx)(x-a) Vertical distance between f and secant line is G(x)= f(x) - f(a) - (Δf/Δx)(x-a) G(x) is continuous on [a, b], differentiable on (a, b), G(a)=G(b), thus by Rolle s theorem there must be a point c in (a, b) such that G (c) = 0. Now G (x)= f (x) - (Δf/Δx), so G (c)= f (c) - (Δf/Δx) =0 hence f (c) = (Δf/Δx). Ͱ
27 MVT is useful to construct proofs Example: Show that sin x < x for all x >0.
28 MVT is useful to construct proofs Example: Show that sin x < x for all x >0. Solution: If x > 2π, then sin x 1 < 2π < x. If 0 < x 2π, then by MVT there exists c in (0, x) such that [sin x sin (0)]/(x-0) = cos(c) < 1 (sin x)/x < 1.
29 MVT allows us to prove other theorems Thrm: If f (x)=0 for all x in (a, b), then f is constant on (a, b).
30 MVT allows us to prove other theorems Thrm: If f (x)=0 for all x in (a, b), then f is constant on (a, b). Proof: By MVT there is c in (a,x) s.t. [f(x) f(a)] /(x-a) = f (c) = 0 so f(x) = f(a) for all x in (a, b).
31 MVT allows us to prove other theorems Thrm: If f (x)=0 for all x in (a, b), then f is constant on (a, b). Proof: By MVT there is c in (a,x) s.t. [f(x) f(a)] /(x-a) = f (c) = 0 so f(x) = f(a) for all x in (a, b). Thrm: If f (x)=g (x) for all x in (a, b), then f-g is constant on (a,b), that is f =g + C, where C is some constant.
32 MVT allows us to prove other theorems Thrm: If f (x)=0 for all x in (a, b), then f is constant on (a, b). Proof: By MVT there is c in (a,x) s.t. [f(x) f(a)] /(x-a) = f (c) = 0 so f(x) = f(a) for all x in (a, b). Thrm: If f (x)=g (x) for all x in (a, b), then f-g is constant on (a,b), that is f =g + C, where C is some constant. Proof: G(x)= f(x)-g(x).by MVT, there is c in (a,x) s.t. [G(x)-G(a)]/(x-a) = G (c) = f (c) g (c) = 0 so G(x)-G(a)=0, hence G(x)=G(a) for all x in (a,b), that is G(x)= f(x)-g(x) = constant.
33 Recap from Tuesday Extreme Value Theorem: If f is cont. on [a,b], f attains max/min values. Fermat s Theorem: If f has a local max or min at x=c, then f (c) = 0. Rolle s Theorem: If f is cont. on [a,b] and diff. on (a,b) and f(a)=f(b), then there exists a number c in (a,b) s.t. f (c)=0 Mean Value Theorem: If f is cont. on [a,b] and diff. on (a,b), then there exists a number c in (a,b) s.t. f(b) f(a)=f (c)(b a)
34 Now we can answer our earlier question Suppose we know f (a) >0, positive slope, increasing f. Can we claim f (x) >0 for all x close to a, i.e. can we claim f is increasing on an interval containing a?
35 Increasing/Decreasing Test If f (x) > 0 on an interval I, then f is increasing for all x in I. If f (x) < 0 on an interval I, then f is decreasing for all x in I. Proof: Hint: Use MVT. Terminology: f is increasing if f (x 1 ) < f (x 2 ) when x 1 < x 2. f is decreasing if f (x 1 ) > f (x 2 ) when x 1 > x 2.
36 Critical points: candidate points for local extrema x=c is a critical number of f is f (c) is zero f (c) is undefined If x=c is a critical number, (c, f(c)) is called critical point
37 First Derivative Test: How to locate local extrema If f (c)=0 or f (c) does not exist (c = critical number of f ) if f changes from + to at c, then f(c) is local max if f changes from to + at c, then f(c) is local min x=c x=c
38 Examples Find all maxima and minima of a) f(x)= x 1/3 (x+4) b) g(x)= x(6-x) 1/2 c) h(x)= (x 2 +4) e x
39 What does f say about f? Concavity Some definitions: f is concave up if at any point on an interval I the graph of f lies above the tangent line. f is concave down if at any point on an interval I the graph of f lies below the tangent line.
40 Concavity Test If f (x) > 0 on an interval I, then the graph of f is concave up for all x in I. If f (x) < 0 on an interval I, then the graph of f is concave down for all x in I. Proof : we need to prove f(x) > f(a) +f (a)(x-a)
41 Inflection points Def: A point P(a, f (a) ) on a curve y = f(x) is an inflection point if f is continuous at x=a and concavity changes sign as f goes through P.
42 Examples Find the concavity and inflection points of a) f(x)= x 1/3 (x+4) b) g(x)= x(6-x) 1/2 c) h(x)= (x 2 +4) e x
43 Second derivative test for extrema If f (c) = 0 and f is continuous on an interval containing c, if f (c) < 0 then f is a local maximum if f (c) > 0 then f is a local minimum Does this mean that if f (a) = 0 there must be an inflection point at x =a?
44 Second derivative test for extrema If f (c) = 0 and f is continuous on an interval containing c, if f (c) < 0 then f is a local maximum if f (c) > 0 then f is a local minimum Does this mean that if f (a) = 0 there must be an inflection point at x =a? No!
45 Problem: Imagine you are skydiving. The graph of your speed as a function of time from the time you jumped out of the plane to the time you achieved terminal velocity is a) increasing and concave up b) increasing and concave down c) decreasing and concave up d) decreasing and concave up
46 If an object of mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account, is v(t) = (mg/c)[1 e (-ct/m) ] Where g is the acceleration due to gravity and c is a positive constant. Verify v(t) is increasing and concave down everywhere.
47 Problem: Suppose f and g are both concave up for all x. Under what condition on f will the composite function h(x)= f(g(x)) be concave up? Assume f and g exist everywhere and they are never zero. Assuming f satisfies the condition above, show that if g has a minimum at x=c, h(x) has also a minimum at x=c.
48 True/False: Let f be a twice differentiable function everywhere, then f is maximum at a point of inflection of f.
49 True/False: Let f be a twice differentiable function everywhere, then f is maximum at a point of inflection of f. False! Let f be a twice differentiable function everywhere, then f has an extreme value at a point of inflection of f, not necessary a maximum (it could be a minimum).
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