# x 2 x 2 cos 1 x x2, lim 1. If x > 0, multiply all three parts by x > 0, we get: x x cos 1 x x, lim lim x cos 1 lim = 5 lim sin 5x

Save this PDF as:

Size: px
Start display at page:

Download "x 2 x 2 cos 1 x x2, lim 1. If x > 0, multiply all three parts by x > 0, we get: x x cos 1 x x, lim lim x cos 1 lim = 5 lim sin 5x"

## Transcription

1 Homework 4 3.4,. Show that x x cos x x holds for x 0. Solution: Since cos x, multiply all three parts by x > 0, we get: x x cos x x, and since x 0 x x 0 ( x ) = 0, then by Sandwich theorem, we get: x 0 x cos x = 0. b. Use the sandwich theorem to show that x 0 x cos x = 0. Solution: Since cos x. If x > 0, multiply all three parts by x > 0, we get: x x cos x x, and since x 0+ x x 0+ ( x) = 0, then by Sandwich theorem, we get: If x < 0,multiply all three parts by x < 0, we get: x cos x 0+ x = 0. x x cos x x, and since x 0 x x 0 ( x) = 0, then by Sandwich theorem, we get: x 0 x cos x = 0. Therefore, we get x cos x 0 x = 0. sin 5x sin 5x 5 x 0 x x 0 5x = 5 sin 5x x 0 5x = 5. sin x x 0 x = sin x x 0 x =. 3.5, 5. Use the intermediate-value theorem to show that e x = x has a solution in (0, ). Solution: Let f(x) = e x x, then f(0) = > 0, and f() = e < 0, thus by I-V theorem, there is c in (0, ), such that f(c) = 0, that is e c = c. 6. Use the intermediate-value theorem to show that cos x = x has a solution in (0, ). Solution: Let f(x) = cos x x, then f(0) = > 0, and f() = cos < 0, thus by I-V theorem, there is c in (0, ), such that f(c) = 0, that is cos c = c. 3 Explain why a polynomial of degree 3 has at least one root. Solution: Suppose f(x) = ax 3 + bx + cx + d is a polynomial of degree 3, where a, b, c, d are constants. If a > 0 then, there is a number t 0 big enough, such that f(t) > 0, and there is a number s 0 small

2 enough, such that f(s) < 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. Similarly, if a < 0 then, there is a number t 0 big enough, such that f(t) < 0, and there is a number s 0 small enough, such that f(s) > 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. 4 Explain why a polynomial of degree n, where n is an odd number has at least one root. Solution: Suppose f(x) = a n x n + a n x n + + a x + a x + a 0 is a polynomial of degree n, where a n, a n,, a 0 are constants. Because n is odd, if a n > 0 then, there is a number t 0 big enough, such that f(t) > 0, and there is a number s 0 small enough, such that f(s) < 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. Similarly, if a n < 0 then, there is a number t 0 big enough, such that f(t) < 0, and there is a number s 0 small enough, such that f(s) > 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. 5. Explain why y = x 4 has at least two roots. Solution: Since y(3) = 5 > 0, and y() = 3 < 0, then by I-V theorem, there is a number c in (, 3), such that y(c ) = c 4 = 0. Also since y( 3) = 5 > 0, and y( ) = 3 < 0, then by I-V theorem, there is a number c in ( 3, ), such that y(c ) = c 4 = 0, and because c and c are in disjoint sets respectively, so they are different, therefore y = x 4 has at least roots c, c. 4., 5. Use the formal definition to find the derivative of y = x, for x > 0. Solution: Denote f(x) = x, then by the formal definition of derivative, for x > 0, we have f f(x + h) f(x) x + h x ( x + h x)( x + h + x) (x) h h h( x + h + x) x + h x h( x + h + x) = x + h + x x. 6. Use the formal definition to find the derivative of f(x) = x+ for x. Solution: Denote f(x) = x, then by the formal definition of derivative, for x, we have f f(x + h) f(x) (x) h h h(x + + h)(x + ) x++h x+ x+ (x++h) (x++h)(x+) h h (x + + h)(x + ) = (x + ). 7. Find the equation of the tangent line to the curve y = 3x at the point (, 3). Solution: y (x) = 6x, then y () = 6, which is the slope of the tangent line. Since the tangent line pass the point (, 3), thus the equation of the tangent line is: y 3 = 6(x ), that is y = 6x Find the equation of the tangent line to the curve y = /x at the point (, ). Solution: y (x) = x, then y () =, which is the slope of the tangent line. Since the tangent line pass the point (, ), thus the equation of the tangent line is: y = (x ), that is y = x Which of the following statements is true: (A) If f(x) is continuous, then f(x) is differentiable. (B) If f(x) is differentiable, then f(x) is continuous. Solution: (A) is false. Counterexample: f(x) = x /3 is continuous at x = 0, but not differentiable at x = 0. (B) is true, the proof is in page 4-4.

3 5.Explain the relation between continuity and differentability. Solution: If f is differentiable at x = c, then f is also continuous at x = c; but if f is continuous at x = c, then f need not be differentiable at x = c. 4., 7. Differentiate the function g(s) = 5s 7 + s 3 5s. Solution: g (s) = 5 7s 6 + 3s 5 = 35s 6 + 6s 5. 8.Differentiate the function g(s) = 3 4s 4s 3. Solution: g (s) = 4 s 4 3s = 8s s. 9. Differentiate f(x) = x 3 rx s with respect to x. Assume that r and s are constant. Solution:f (s) = 3x r = 3 x r. 30 Differentiate f(x) = r+x x + (r + s)x with respect to x. Assume that r and s are nonzero constants. Solution: f (x) = ( r + x x + (r + s)x) = + r + s. 8. Suppose that P (x) is a polynomial of degree 4. Is P (x) a polynomial as well? If yes, what is the degree. Solution: Yes. Because if P (x) = ax 4 + bx 3 + cx + dx + e, where a, b, c, d, e are all constants. then P (x) = 4ax 3 + 3bx + cx + d, therefore degree of P (x) is Suppose that P (x) is a polynomial of degree k. Is P (x) a polynomial as well? If yes, what is the degree? Solution: Suppose P (x) = a n x n + a n x n + + a x + a x + a 0, then P (x) = na n x n + (n )a n x n + + a x + a, thus the degree of P (x) is n. 4.3, 3. Differentiate f(x) = a(x a) + a with respect to x. Assume that a is a positive constant. Solution: Use the product rule to f(x) = a(x a)(x a) + a: f (x) = a x (x ) + a x (x ) = 8ax(x a) = 8ax 3 8a x. 3 Differentiate f(x) = 3(x ) +a with respect to x. Assume that a is a positive constant. Solution: Use the product rule to f(x) = 3 +a (x )(x ), we get: Differentiate the following functions: 67.g(s) = s/3 s /3. Solution: g (s) = 68.gs = s/7 s /7 s 3/7 +s 4/7. Solution: f (x) = 3 6 ((x ) + (x )) = (x ) + a s /3 (s /3 ) 3 (s/3 )s /3 (s /3 ) = 3 3 s / s /3 (s /3 ) = 3 3 s /3 + 3 s /3 (s /3 ). g (s) = ( 7 s 6/7 7 s 5/7 )(s 3/7 + s 4/7 ) (s /7 s /7 )( 3 7 s 4/ s 3/7 ) (s 3/7 + s 4/7 ) = 7 s 3/7 s /7 + s /7 (s 3/7 + s 4/7 ) 3

4 Assume that f(x) is differentiable. Find an expression for the derivative of y at x =, assuming that f() = and f () = : 84. y = f(x) x +. Solution: y (x) = f (x)(x +) xf(x) (x +), then: y () = f ()( + ) 4f() ( + ) = = y = x +4f(x) f(x) Solution:y (x) = (x+4f (x))f(x) (x +4f(x))f (x) (f(x) ), thus: y () = (4 + 4f ())f() ( + 4f())f () (f() ) = 8 ( ) 0 = 8. 4

### Solutions to Homework 6 Mathematics 503 Foundations of Mathematics Spring 2014

Solutions to Homework 6 Mathematics 503 Foundations of Mathematics Spring 2014 3.4: 1. If m is any integer, then m(m + 1) = m 2 + m is the product of m and its successor. That it to say, m 2 + m is the

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### Critical points of once continuously differentiable functions are important because they are the only points that can be local maxima or minima.

Lecture 0: Convexity and Optimization We say that if f is a once continuously differentiable function on an interval I, and x is a point in the interior of I that x is a critical point of f if f (x) =

### Approximating functions by Taylor Polynomials.

Chapter 4 Approximating functions by Taylor Polynomials. 4.1 Linear Approximations We have already seen how to approximate a function using its tangent line. This was the key idea in Euler s method. If

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### 1. Graphing Linear Inequalities

Notation. CHAPTER 4 Linear Programming 1. Graphing Linear Inequalities x apple y means x is less than or equal to y. x y means x is greater than or equal to y. x < y means x is less than y. x > y means

### (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order 4; (1, 0) : order 2; (1, 1) : order 4; (1, 2) : order 2; (1, 3) : order 4.

11.01 List the elements of Z 2 Z 4. Find the order of each of the elements is this group cyclic? Solution: The elements of Z 2 Z 4 are: (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order

### x 2 if 2 x < 0 4 x if 2 x 6

Piecewise-defined Functions Example Consider the function f defined by x if x < 0 f (x) = x if 0 x < 4 x if x 6 Piecewise-defined Functions Example Consider the function f defined by x if x < 0 f (x) =

### (LMCS, p. 317) V.1. First Order Logic. This is the most powerful, most expressive logic that we will examine.

(LMCS, p. 317) V.1 First Order Logic This is the most powerful, most expressive logic that we will examine. Our version of first-order logic will use the following symbols: variables connectives (,,,,

### Introduction to Differential Calculus. Christopher Thomas

Mathematics Learning Centre Introduction to Differential Calculus Christopher Thomas c 1997 University of Sydney Acknowledgements Some parts of this booklet appeared in a similar form in the booklet Review

### Properties of BMO functions whose reciprocals are also BMO

Properties of BMO functions whose reciprocals are also BMO R. L. Johnson and C. J. Neugebauer The main result says that a non-negative BMO-function w, whose reciprocal is also in BMO, belongs to p> A p,and

### Preliminary Version: December 1998

On the Number of Prime Numbers less than a Given Quantity. (Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse.) Bernhard Riemann [Monatsberichte der Berliner Akademie, November 859.] Translated

### 2 Polynomials over a field

2 Polynomials over a field A polynomial over a field F is a sequence (a 0, a 1, a 2,, a n, ) where a i F i with a i = 0 from some point on a i is called the i th coefficient of f We define three special

### WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?

WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT? introduction Many students seem to have trouble with the notion of a mathematical proof. People that come to a course like Math 216, who certainly

### 3. Which of the following couldn t be the solution of a differential equation? (a) z(t) = 6

MathQuest: Differential Equations What is a Differential Equation? 1. Which of the following is not a differential equation? (a) y = 3y (b) 2x 2 y + y 2 = 6 (c) tx dx dt = 2 (d) d2 y + 4 dy dx 2 dx + 7y

### You know from calculus that functions play a fundamental role in mathematics.

CHPTER 12 Functions You know from calculus that functions play a fundamental role in mathematics. You likely view a function as a kind of formula that describes a relationship between two (or more) quantities.

### Inversion. Chapter 7. 7.1 Constructing The Inverse of a Point: If P is inside the circle of inversion: (See Figure 7.1)

Chapter 7 Inversion Goal: In this chapter we define inversion, give constructions for inverses of points both inside and outside the circle of inversion, and show how inversion could be done using Geometer

### Math 2001 Homework #10 Solutions

Math 00 Homework #0 Solutions. Section.: ab. For each map below, determine the number of southerly paths from point to point. Solution: We just have to use the same process as we did in building Pascal

### MUST-HAVE MATH TOOLS FOR GRADUATE STUDY IN ECONOMICS

MUST-HAVE MATH TOOLS FOR GRADUATE STUDY IN ECONOMICS William Neilson Department of Economics University of Tennessee Knoxville September 29 28-9 by William Neilson web.utk.edu/~wneilson/mathbook.pdf Acknowledgments

### This document has been written by Michaël Baudin from the Scilab Consortium. December 2010 The Scilab Consortium Digiteo. All rights reserved.

SCILAB IS NOT NAIVE This document has been written by Michaël Baudin from the Scilab Consortium. December 2010 The Scilab Consortium Digiteo. All rights reserved. December 2010 Abstract Most of the time,

### Some Basic Techniques of Group Theory

Chapter 5 Some Basic Techniques of Group Theory 5.1 Groups Acting on Sets In this chapter we are going to analyze and classify groups, and, if possible, break down complicated groups into simpler components.

### how to use dual base log log slide rules

how to use dual base log log slide rules by Professor Maurice L. Hartung The University of Chicago Pickett The World s Most Accurate Slide Rules Pickett, Inc. Pickett Square Santa Barbara, California 93102

### The Optimality of Naive Bayes

The Optimality of Naive Bayes Harry Zhang Faculty of Computer Science University of New Brunswick Fredericton, New Brunswick, Canada email: hzhang@unbca E3B 5A3 Abstract Naive Bayes is one of the most

### Q D = 100 - (5)(5) = 75 Q S = 50 + (5)(5) = 75.

4. The rent control agency of New York City has found that aggregate demand is Q D = 100-5P. Quantity is measured in tens of thousands of apartments. Price, the average monthly rental rate, is measured

### A Modern Course on Curves and Surfaces. Richard S. Palais

A Modern Course on Curves and Surfaces Richard S. Palais Contents Lecture 1. Introduction 1 Lecture 2. What is Geometry 4 Lecture 3. Geometry of Inner-Product Spaces 7 Lecture 4. Linear Maps and the Euclidean

### Space-Time Approach to Non-Relativistic Quantum Mechanics

R. P. Feynman, Rev. of Mod. Phys., 20, 367 1948 Space-Time Approach to Non-Relativistic Quantum Mechanics R.P. Feynman Cornell University, Ithaca, New York Reprinted in Quantum Electrodynamics, edited

### A 2-factor in which each cycle has long length in claw-free graphs

A -factor in which each cycle has long length in claw-free graphs Roman Čada Shuya Chiba Kiyoshi Yoshimoto 3 Department of Mathematics University of West Bohemia and Institute of Theoretical Computer Science

### If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?

Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question