MTH4100 Calculus I. Bill Jackson School of Mathematical Sciences QMUL. Week 9, Semester 1, 2013

Transcription

1 MTH4100 School of Mathematical Sciences QMUL Week 9, Semester 1, 2013

2 Concavity

3 Concavity In the literature concave up is often referred to as convex, and concave down is simply called concave.

4 The second derivative test for concavity If f is twice differential on an interval I, the First Derivative Test for Monotonic Functions implies that f increases on I if f (x) > 0 for all x I and decreases if f (x) < 0 for all x I. This gives:

5 The second derivative test for concavity If f is twice differential on an interval I, the First Derivative Test for Monotonic Functions implies that f increases on I if f (x) > 0 for all x I and decreases if f (x) < 0 for all x I. This gives: This tells us that a point c at which the function f changes from being concave up to concave down must be a critical point of f i.e. we have either f (c) = 0 or f (c) undefined.

6 The second derivative test for concavity If f is twice differential on an interval I, the First Derivative Test for Monotonic Functions implies that f increases on I if f (x) > 0 for all x I and decreases if f (x) < 0 for all x I. This gives: This tells us that a point c at which the function f changes from being concave up to concave down must be a critical point of f i.e. we have either f (c) = 0 or f (c) undefined. Example Find the intervals on the real line for which the graphs of the following functions are concave up or concave down: (1) y = x 3 (2) y = x 1/3

7 Points of inflection

8 Points of inflection Examples: y = x 3 ; y = x 3 3x.

9 Points of inflection Examples: y = x 3 ; y = x 3 3x. The condition that the graph of the function has a tangent line at a point of inflection is slightly more general than saying that the function is differentiable at the point since it allows the tangent line to be vertical (and hence the derivative to be infinite ).

10 Points of inflection Examples: y = x 3 ; y = x 3 3x. The condition that the graph of the function has a tangent line at a point of inflection is slightly more general than saying that the function is differentiable at the point since it allows the tangent line to be vertical (and hence the derivative to be infinite ). Example: y = x 1/3.

11 Points of inflection Note that for (c,f(c)) to be a point of inflection of f we do not need to have f (c) = 0. Example y = x 3 3x.

12 Points of inflection Note that for (c,f(c)) to be a point of inflection of f we do not need to have f (c) = 0. Example y = x 3 3x. Note that we can have f (c) = 0 WITHOUT (c,f(c)) being a point of inflection. Example y = x 4.

13 Second derivative test for local extrema When f is a continuous function, c is a critical point of f, and f is twice differentiable at c, the second derivative f (c) often determines whether f(c) is a local maximum or minimum of f:

14 Second derivative test for local extrema When f is a continuous function, c is a critical point of f, and f is twice differentiable at c, the second derivative f (c) often determines whether f(c) is a local maximum or minimum of f: Theorem (Second derivative test for local extrema) Suppose f is a function, f (c) = 0, and f is continuous on some open interval which contains c. 1 If f (c) < 0 then f has a local maximum at c. 2 If f (c) > 0 then f has a local minimum at c. 3 If f (c) = 0 then the test fails, f can have either a local maximum, a local minimum, or a point of inflection at c.

15 Second derivative test for local extrema When f is a continuous function, c is a critical point of f, and f is twice differentiable at c, the second derivative f (c) often determines whether f(c) is a local maximum or minimum of f: Theorem (Second derivative test for local extrema) Suppose f is a function, f (c) = 0, and f is continuous on some open interval which contains c. 1 If f (c) < 0 then f has a local maximum at c. 2 If f (c) > 0 then f has a local minimum at c. 3 If f (c) = 0 then the test fails, f can have either a local maximum, a local minimum, or a point of inflection at c. Note. In case (3) we can use the first derivative test to determine if f has a local maximum or minimum or a point of inflection at c i.e. check whether f (x) changes sign at x = c.

16 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have.

17 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have. Step 2 Determine f and f.

18 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have. Step 2 Determine f and f. Step 3 Find the critical points of f and determine the functions behavior at each one.

19 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have. Step 2 Determine f and f. Step 3 Find the critical points of f and determine the functions behavior at each one. Step 4 Determine where f is increasing or decreasing.

20 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have. Step 2 Determine f and f. Step 3 Find the critical points of f and determine the functions behavior at each one. Step 4 Determine where f is increasing or decreasing. Step 5 Find the points of inflection, if any occur, and determine where the graph is concave up or concave down.

21 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have. Step 2 Determine f and f. Step 3 Find the critical points of f and determine the functions behavior at each one. Step 4 Determine where f is increasing or decreasing. Step 5 Find the points of inflection, if any occur, and determine where the graph is concave up or concave down. Step 6 Investigate the behavior of f(x) as x ± and identify any asymptotes.

22 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have. Step 2 Determine f and f. Step 3 Find the critical points of f and determine the functions behavior at each one. Step 4 Determine where f is increasing or decreasing. Step 5 Find the points of inflection, if any occur, and determine where the graph is concave up or concave down. Step 6 Investigate the behavior of f(x) as x ± and identify any asymptotes. Step 7 Plot key points such as the intercepts of the graph on the axes and the points found in Steps 3-5, then sketch the graph.

23 Graph Drawing: Strategy for Graphing y = f(x) Step 1 Identify the natural domain of f and find any symmetries the graph may have. Step 2 Determine f and f. Step 3 Find the critical points of f and determine the functions behavior at each one. Step 4 Determine where f is increasing or decreasing. Step 5 Find the points of inflection, if any occur, and determine where the graph is concave up or concave down. Step 6 Investigate the behavior of f(x) as x ± and identify any asymptotes. Step 7 Plot key points such as the intercepts of the graph on the axes and the points found in Steps 3-5, then sketch the graph. Example Sketch the graph of f(x) = (x +1)2 1+x 2.

24 Summary: Learning about functions from derivatives

25 L Hôpital s Rule and Indeterminate Forms If f(a) = g(a) = 0, then f(a)/g(a) = 0/0 is a meaningless expression, called an indeterminate form.

26 L Hôpital s Rule and Indeterminate Forms If f(a) = g(a) = 0, then f(a)/g(a) = 0/0 is a meaningless expression, called an indeterminate form. f(x) In this case lim cannot be found by simply substituting x a g(x) x = a.

27 L Hôpital s Rule and Indeterminate Forms If f(a) = g(a) = 0, then f(a)/g(a) = 0/0 is a meaningless expression, called an indeterminate form. f(x) In this case lim cannot be found by simply substituting x a g(x) x = a. L Hôpital s Rule gives us a method to calculate this limit if f and g are both differentiable at x = a.

28 L Hôpital s Rule and Indeterminate Forms If f(a) = g(a) = 0, then f(a)/g(a) = 0/0 is a meaningless expression, called an indeterminate form. f(x) In this case lim cannot be found by simply substituting x a g(x) x = a. L Hôpital s Rule gives us a method to calculate this limit if f and g are both differentiable at x = a. Theorem (L Hôpital s Rule) Suppose that f(a) = g(a) = 0, that f and g are differentiable on an open interval I containing a, and that g (x) 0 if x a. Then f(x) lim x a g(x) = lim f (x) x a g (x), whenever the limit on the right side exists.

29 L Hôpital s Rule and Indeterminate Forms If f(a) = g(a) = 0, then f(a)/g(a) = 0/0 is a meaningless expression, called an indeterminate form. f(x) In this case lim cannot be found by simply substituting x a g(x) x = a. L Hôpital s Rule gives us a method to calculate this limit if f and g are both differentiable at x = a. Theorem (L Hôpital s Rule) Suppose that f(a) = g(a) = 0, that f and g are differentiable on an open interval I containing a, and that g (x) 0 if x a. Then f(x) lim x a g(x) = lim f (x) x a g (x), whenever the limit on the right side exists. Example: Determine lim x 0 5x sinx x

30 L Hôpital s Rule - Notes Note 1 Always check f(a) = g(a) = 0, before you try to use f(x) l Hôpital to calculate lim x a g(x). Otherwise you may get a wrong answer: 1+sinx Example: lim x 0 1 x.

31 L Hôpital s Rule - Notes Note 1 Always check f(a) = g(a) = 0, before you try to use f(x) l Hôpital to calculate lim x a g(x). Otherwise you may get a wrong answer: 1+sinx Example: lim x 0 1 x. Note 2 Sometimes we have to use l Hôpital s rule recursively: Example: Determine x sinx lim x 0 x 3.

32 L Hôpital s Rule - Notes Note 1 Always check f(a) = g(a) = 0, before you try to use f(x) l Hôpital to calculate lim x a g(x). Otherwise you may get a wrong answer: 1+sinx Example: lim x 0 1 x. Note 2 Sometimes we have to use l Hôpital s rule recursively: Example: Determine x sinx lim x 0 x 3. Note 3 L Hôpital s rule can also be applied to one-sided limits. Example: sinx lim x 0 + x 2

33 Other indeterminate forms L Hôpital s rule can also be applied to other indeterminate forms such as /, 0 and.

34 Other indeterminate forms L Hôpital s rule can also be applied to other indeterminate forms such as /, 0 and. / : If lim x a f(x) = = lim x a g(x), then use f(x) lim x a g(x) = lim x a f (x) g (x). x x Example: Determine lim 2 x x 2 +7x.

35 Other indeterminate forms L Hôpital s rule can also be applied to other indeterminate forms such as /, 0 and. / : If lim x a f(x) = = lim x a g(x), then use f(x) lim x a g(x) = lim x a f (x) g (x). x x Example: Determine lim 2 x x 2 +7x. 0 : If lim x a f(x) = and lim x a g(x) = 0, then lim x a 1/f(x) = 0 and we can use g(x) lim x a (f(x)g(x)) = lim x a 1/f(x). Example: Determine lim x x sin(1/x).

36 Other indeterminate forms L Hôpital s rule can also be applied to other indeterminate forms such as /, 0 and. / : If lim x a f(x) = = lim x a g(x), then use f(x) lim x a g(x) = lim x a f (x) g (x). x x Example: Determine lim 2 x x 2 +7x. 0 : If lim x a f(x) = and lim x a g(x) = 0, then lim x a 1/f(x) = 0 and we can use g(x) lim x a (f(x)g(x)) = lim x a 1/f(x). Example: Determine lim x x sin(1/x). : Try to gather terms so we can use the standard form of L Hôpital rule: Example: Determine lim x 0 ( 1 sinx 1 x).

37 Antiderivatives Idea: Given a function f, find a function F such that F = f.

38 Antiderivatives Idea: Given a function f, find a function F such that F = f.

39 Antiderivatives Idea: Given a function f, find a function F such that F = f. Examples: f(x) = 2x, h(x) = sinx.

40 General antiderivatives It is easy to see that if F(x) is an antiderivative of f(x) then F(x)+C will be an antiderivative of f(x) for any constant C R.

41 General antiderivatives It is easy to see that if F(x) is an antiderivative of f(x) then F(x)+C will be an antiderivative of f(x) for any constant C R. Furthermore, if G(x) is any other antiderivative of f(x) then we have F (x) = f(x) = G (x) and the second corollary to the Mean Value Theorem tells us that G(x) = F(x)+C for some constant C R. This gives:

42 General antiderivatives It is easy to see that if F(x) is an antiderivative of f(x) then F(x)+C will be an antiderivative of f(x) for any constant C R. Furthermore, if G(x) is any other antiderivative of f(x) then we have F (x) = f(x) = G (x) and the second corollary to the Mean Value Theorem tells us that G(x) = F(x)+C for some constant C R. This gives:

43 Some common antideratives These formula can easily be verified by showing that the derivative of each antiderivative is equal to the given function.

44 Antiderivative linearity rules Lemma Suppose f(x),g(x) are functions with antiderivatives F(x) and G(x), and k R. Then: kf(x) has general antiderivative kf(x)+c; f(x)+g(x) has general antiderivative F(x)+G(x)+C; for an arbitrary constant C R.

45 Antiderivative linearity rules Lemma Suppose f(x),g(x) are functions with antiderivatives F(x) and G(x), and k R. Then: kf(x) has general antiderivative kf(x)+c; f(x)+g(x) has general antiderivative F(x)+G(x)+C; for an arbitrary constant C R. Example: Find the general antiderivative of h(x) = 5 x +sin3x.

46 Notation for antiderivatives Definition We refer to the general antiderivative, F(x)+C, of f(x) as the indefinite integral of f(x) and denote it by f(x)dx. The symbol is called an integral sign, the function f is the integrand and the variable x is the variable of integration.

47 Notation for antiderivatives Definition We refer to the general antiderivative, F(x)+C, of f(x) as the indefinite integral of f(x) and denote it by f(x)dx. The symbol is called an integral sign, the function f is the integrand and the variable x is the variable of integration. Examples: 4x dx, cosx dx

48 Estimating areas with finite sums Example: How can we compute the area of the shaded region R?

49 Approximation algorithm Subdivide the interval [0, 1] into n subintervals of equal width x = 1 n. Choose a point c k in the k th subinterval. For example we could use: 1 midpoint rule: Choose c k in the middle of the k th subinterval. 2 max rule: Choose c k such that f(c k )is maximum. 3 min rule: choose c k such that f(c k ) is minimum.

50 Approximation algorithm - continued Construct n rectangles with base x and height c k. Approximate the area by calculating the sum of the areas of these rectangles f(c 1 ) x +f(c 2 ) x +...+f(c n ) x.

51 Approximation algorithm - continued Construct n rectangles with base x and height c k. Approximate the area by calculating the sum of the areas of these rectangles f(c 1 ) x +f(c 2 ) x +...+f(c n ) x. Note that the area of R will lie between the upper sum i.e the sum we obtain using the max rule to choose the points c k and the lower sum i.e the sum we obtain using the min rule to choose the points c k. So we can estimate how close our approximation is to the correct area by calculating the difference between these two sums.

52 Approximation algorithm - continued We can improve the accuracy of our approximation by choosing shorter subintervals i.e. larger values for n.

53 Approximation algorithm - continued We can improve the accuracy of our approximation by choosing shorter subintervals i.e. larger values for n. If the approximations converge to the same limit as n, no matter how we choose the points c k, then this limit will be the area of R.

54 Finite sums To handle sums with many terms, we need a more concise notation. Let where n a k = a 1 +a a n k=1 Example: 3 k=1 ( 1)k k

55 Algebraic properties of sums Example: by rules 1 and 2. n n (5k k 3 ) = 5 k k=1 n k=1 k=1 k 3

56 Some common sums Theorem Sum of first n natural numbers: Sum of first n squares: n k=1 k = n(n+1). 2 n k 2 = k=1 n(n+1)(2n Sum of first n cubes: n ( ) n(n+1) 2 k 3 =. 2 k=1

57 Example continued We can now compute the area of the region R below the graph of y = 1 x 2 and above the interval [0,1].

58 Example continued We can now compute the area of the region R below the graph of y = 1 x 2 and above the interval [0,1]. [ Subdivide [ the interval into n subintervals of width x = 1 n : 0, 1 n], 1 n, n] 2 [, 2 n, 3 ] [ n,..., n 1 n, n n].

59 Example continued We can now compute the area of the region R below the graph of y = 1 x 2 and above the interval [0,1]. [ Subdivide [ the interval into n subintervals of width x = 1 n : 0, 1 n], 1 n, n] 2 [, 2 n, 3 ] [ n,..., n 1 n, n n]. Use the min rule to choose the points c k : this gives c k = k n, k N is the rightmost point in the k th subinterval.

60 Example continued We can now compute the area of the region R below the graph of y = 1 x 2 and above the interval [0,1]. [ Subdivide [ the interval into n subintervals of width x = 1 n : 0, 1 n], 1 n, n] 2 [, 2 n, 3 ] [ n,..., n 1 n, n n]. Use the min rule to choose the points c k : this gives c k = k n, k N is the rightmost point in the k th subinterval. The lower sum is f ( ) 1 1 n n +f ( ) 2 1 n n +...+f ( ) n 1 n n = n k=1 f ( ) k 1 n n = n 1. 6n 2 Hence the area of R is at least n 1 6n 2.

61 Example continued We can now compute the area of the region R below the graph of y = 1 x 2 and above the interval [0,1]. [ Subdivide [ the interval into n subintervals of width x = 1 n : 0, 1 n], 1 n, n] 2 [, 2 n, 3 ] [ n,..., n 1 n, n n]. Use the min rule to choose the points c k : this gives c k = k n, k N is the rightmost point in the k th subinterval. The lower sum is f ( ) 1 1 n n +f ( ) 2 1 n n +...+f ( ) n 1 n n = n k=1 f ( ) k 1 n n = n 1. 6n 2 Hence the area of R is at least n 1. 6n 2 A similar calculation shows that the upper sum is n 1 6n 2 and hence the area of R is at most n 1. 6n 2

62 Example continued We can now compute the area of the region R below the graph of y = 1 x 2 and above the interval [0,1]. [ Subdivide [ the interval into n subintervals of width x = 1 n : 0, 1 n], 1 n, n] 2 [, 2 n, 3 ] [ n,..., n 1 n, n n]. Use the min rule to choose the points c k : this gives c k = k n, k N is the rightmost point in the k th subinterval. The lower sum is f ( ) 1 1 n n +f ( ) 2 1 n n +...+f ( ) n 1 n n = n k=1 f ( ) k 1 n n = n 1. 6n 2 Hence the area of R is at least n 1. 6n 2 A similar calculation shows that the upper sum is n 1 6n 2 and hence the area of R is at most n 1. 6n 2 As n, both sums converge to 2 3. Therefore, the area of R is 2 3.

63 Example continued We can now compute the area of the region R below the graph of y = 1 x 2 and above the interval [0,1]. [ Subdivide [ the interval into n subintervals of width x = 1 n : 0, 1 n], 1 n, n] 2 [, 2 n, 3 ] [ n,..., n 1 n, n n]. Use the min rule to choose the points c k : this gives c k = k n, k N is the rightmost point in the k th subinterval. The lower sum is f ( ) 1 1 n n +f ( ) 2 1 n n +...+f ( ) n 1 n n = n k=1 f ( ) k 1 n n = n 1. 6n 2 Hence the area of R is at least n 1. 6n 2 A similar calculation shows that the upper sum is n 1 6n 2 and hence the area of R is at most n 1. 6n 2 As n, both sums converge to 2 3. Therefore, the area of R is 2 3. Note that any other choice of c k would give the same limit (since its sum must lie between the upper and lower sums).