MA 137: Calculus I for the Life Sciences

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1 MA 137: Calculus I for the Life Sciences David Murrugarra Department of Mathematics, University of Kentucky Spring 2018 David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

2 Topics covered during last class Limits at infinity, Applications, The Intermediate Value Theorem (IVT). What are we going to learn today? Review of the Intermediate Value Theorem (IVT), The Bisection Method, Applications. The Sandwich (Squeeze) Theorem. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

3 Theorem (The Intermediate Value Theorem (B. Bolzano, 1817)) Suppose that f is continuous on the closed interval [a, b]. If u is any real number with f (a) < u < f (b) [or f (b) < u < f (a)], then there exists at least one value c (a, b) such that f (c) = u. y ƒ(b) y = ƒ(x) y = u ƒ(a) a c b x David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

4 Section 3.3: The Intermediate Value Theorem (IVT) In applying the Intermediate Value Theorem, it is important to check that f is continuous. Discontinuous functions can easily miss values; for example, the floor function misses all numbers that are not integers. The Intermediate Value Theorem gives us only the existence of a number c; it does not tell us how many such points there are or where they are located. As an application, the Theorem can be used to find approximate roots (or solutions) of equations of the form f (x) = 0. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

5 Section 3.5: The Intermediate Value Theorem (IVT) Example (Online Homework HW09, #9) Determine if the Intermediate Value Theorem implies that the equation x 3 3x 9.9 = 0 has a root in the interval [0, 1]. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

6 Section 3.5: The Intermediate Value Theorem (IVT) Example (Online Homework HW09, #9) Determine if the Intermediate Value Theorem implies that the equation x 3 3x 9.9 = 0 has a root in the interval [0, 1]. Solution: Let f (x) = x 3 3x 9.9. Then f is continuous on [0, 1] because it is a polynomial. Since f (0) = 9.9 and f (1) = 11.9, the IVT does not apply on [0, 1]. We cannot conclude whether there is a root of f on [0, 1]. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

7 Section 3.5: The Intermediate Value Theorem (IVT) Example (Online Homework HW09, #9) Determine if the Intermediate Value Theorem implies that the equation x 3 3x = 0 has a root in the interval [0, 1]. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

8 Section 3.5: The Intermediate Value Theorem (IVT) Example (Online Homework HW09, #9) Determine if the Intermediate Value Theorem implies that the equation x 3 3x = 0 has a root in the interval [0, 1]. Solution: Let f (x) = x 3 3x Then f is continuous on [0, 1] because it is a polynomial. Since f (0) = 1.2 and f (1) = 0.8, the IVT does apply on [0, 1]. By the IVT, we can conclude that there is a root of f on [0, 1]. That is, there is c (0, 1) such that f (c) = 0. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

9 Section 3.5: Bisection Method The bisection method is used for numerically finding a root of the equation f (x) = 0, where f is a continuous function defined on an interval [a 1, b 1 ] and where f (a 1 ) and f (b 1 ) have opposite signs. F(x) At each step the method divides the interval in two by computing the midpoint b 2 = (a 1 + b 1 )/2 of the interval and the value of the function f (c) at that point. F(a 1 ) F(a 2 ) F(a 3 ) Unless b 2 is itself a root (which is very unlikely, but possible) there are only two possibilities: either f (a 1 ) and f (b 2 ) have opposite signs and [a 1, b 2 ] contains a root, or f (b 2 ) and f (b 1 ) have opposite signs and [b 2, b 1 ] contains a root. a 1 F(b 2 ) b 1 x In this way an interval that contains a zero of f is reduced in width by 50% at each step. The process is repeated until the interval is sufficiently small. F(b 1 ) The bisection method. The bigger red dot is the root of the function, By Tokuchan, David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

10 Section 3.5: Bisection Method Example (Online Homework HW09, #10) Carry out three steps of the Bisection Method for f (x) = 2 x x 4 as follows: 1 Show that f (x) has a zero in [1, 2]. 2 Determine which subinterval, [1, 1.5] or [1.5, 2], contains a zero. 3 Determine which interval, [1, 1.25], [1.25, 1.5], [1.5, 1.75], or [1.75, 2], contains a zero. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

11 Section 3.4: The Sandwich (Squeeze) Theorem Example Suppose we want to calculate lim x e x cos(10x). We soon realize that none of the rules we have learned so far apply. Although lim x e x = 0, we find that lim cos(10x) x does not exist as the function cos(10x) oscillates between 1 and 1. We need to employ some other techniques. One of these techniques is to use the Squeeze (Sandwich) Theorem. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

12 Section 3.4: The Sandwich (Squeeze) Theorem Theorem (Sandwich (Squeeze) Theorem) Consider three functions f (x), g(x) and h(x) and suppose for all x in an open interval that contains c (except possibly at c) we have f (x) g(x) h(x). If lim f (x) = L = lim h(x) then lim g(x) = L. x c x c x c David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

13 Section 3.4: The Sandwich (Squeeze) Theorem Example Suppose we want to calculate lim x e x cos(10x). From the inequality 1 cos(10x) 1 it follows that (as e x > 0, always) e x cos(10x) e x So by the Squeeze Theorem it follows that lim x e x cos(10x) = 0. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

14 Section 3.4: The Sandwich (Squeeze) Theorem Example (Online Homework HW10, #2) Suppose Use this to compute 8x 22 f (x) x 2 2x 13. lim f (x) x 3 David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

15 Section 3.4: The Sandwich (Squeeze) Theorem Example (Neuhauser, Example # 1, p. 114) Find ( ) 1 lim x 2 sin x 0 x Figure: The graph of x 2 sin( 1 x ) in red. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

16 Section 3.4: Trigonometric Limits The following two trigonometric limits are important for developing the differential calculus for trigonometric functions: Rule sin x lim x 0 x = 1 and lim x 0 1 cos x x = 0. Note that the angle x is measured in radians. The proof of the first statement uses a nice geometric argument and the sandwich theorem. The second statement follows from the first. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

17 Trigonometric Functions We will sometimes use the double angle formulas 1 cos(2α) = cos 2 (α) sin 2 (α) 2 sin(2α) = 2 sin α cos α. which are special cases of the following addition formulas 1 cos(α + β) = cos β cos α sin α sin β. 2 sin(α + β) = sin α cos β + cos α sin β. What about sin(α/2) and cos(α/2)? 1 + cos α 1 cos α cos(α/2) = ± and sin(α/2) = ± 2 2 (the sign (+ or -) depends on the quadrant in which α/2 lies.) David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

18 Section 3.4: Trigonometric Limits Example (Online Homework HW10, #7) Evaluate sin 4θ sin 8θ lim θ 0 θ 2. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

19 Section 3.4: Trigonometric Limits Example (Online Homework HW10, #10) Evaluate the limit: tan 5x lim x 0 tan 6x David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

20 Section 3.4: Trigonometric Limits Example (Neuhauser, Example 3(c), p. 118) Evaluate the limit: sec x 1 lim x 0 x sec x David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

Example (Online Homework HW10, #14) A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like an ice cream cone, as shown in the figure.

21 Example (Online Homework HW10, #14) A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like an ice cream cone, as shown in the figure. If A(θ) is the area of the semicircle and B(θ) is the area of the triangle, A(θ) find lim θ 0 + B(θ). Figure: Ice cream cone. David Murrugarra (University of Kentucky) MA 137: Lecture 16 Spring / 19

Continuity. MATH 161 Calculus I. J. Robert Buchanan. Fall Department of Mathematics

Continuity. MATH 161 Calculus I. J. Robert Buchanan. Fall Department of Mathematics Continuity MATH 161 Calculus I J. Robert Buchanan Department of Mathematics Fall 2017 Intuitive Idea A process or an item can be described as continuous if it exists without interruption. The mathematical