N5 Mathematics

Mathematics Downloaded from hlayiso.com N5

Module 1: Limits and continuity INTRODUCTION If 𝑓𝑓(𝑥𝑥) is defined for all 𝑥𝑥 near point 𝑎𝑎 and we can take 𝑓𝑓(𝑥𝑥) close to L by taking 𝑥𝑥 arbitrarily close to 𝑎𝑎, then we say that 𝑓𝑓 approaches the limit L as 𝑥𝑥 approaches a. A function is continuous if its graph can be drawn without picking up the pencil, or more formally, function 𝑓𝑓(𝑥𝑥) is continuous if the limit of 𝑓𝑓(𝑥𝑥), as 𝑥𝑥 approaches 𝑎𝑎 from either direction,from Downloaded is equal to 𝑓𝑓(𝑎𝑎).. hlayiso.com www.futuremanagers.com

Module 1: Limits and continuity (continued) L’Hôpital’s rule The L’Hôpital’s rule can be defined as follows: Suppose that the functions 𝑓𝑓 and 𝑔𝑔 are differentiable on the interval (𝑎𝑎; 𝑏𝑏) where 𝑔𝑔𝑔(𝑥𝑥) ≠ 0. Then: 𝑓𝑓 𝑥𝑥 lim = 𝐿𝐿 Downloaded from 𝑥𝑥→𝑏𝑏 𝑔𝑔 𝑥𝑥 hlayiso.com www.futuremanagers.com

Module 1: Limits and continuity (continued) CONTINUITY Continuity at a point is defined when the limit of the function from the left equals the limit from the right and this value is also equal to the value of the function at that particular point i.e. lim 𝑓𝑓 𝑥𝑥 = lim 𝑓𝑓(𝑥𝑥) = 𝑓𝑓(𝑎𝑎) 𝑥𝑥→𝑎𝑎 𝑥𝑥→𝑎𝑎 Downloaded from hlayiso.com www.futuremanagers.com

Module 2: Differentiation INTRODUCTION Differentiation in mathematics measures how rapidly these functions change at any point with respect to one of their variables. Downloaded from hlayiso.com www.futuremanagers.com

Module 2: Differentiation (continued) DIFFERENTIATION RULES AND TECHNIQUES 𝑓𝑓 𝑥𝑥+ℎ −𝑓𝑓 𝑥𝑥 • The derivative of a function f is defined by: 𝑓𝑓 ′ 𝑥𝑥 = lim ℎ→0 ℎ 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑 • Leibniz notation shows: 𝑓𝑓 ′ 𝑥𝑥 = 𝑦𝑦 ′ = = = 𝑓𝑓 𝑥𝑥 = 𝐷𝐷 𝑥𝑥 = 𝐷𝐷𝑥𝑥 𝑦𝑦 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑑𝑑 𝐷𝐷 𝑑𝑑 𝑑𝑑 𝑑𝑑 Downloaded from hlayiso.com www.futuremanagers.com

Module 3: Application of Differentiation NEWTON’S METHOD TO FIND THE APPROXIMATE VALUE OF IRRATIONAL ROOTS OF AN EQUATION 𝑓𝑓 𝑓𝑓𝑥𝑥𝑛𝑛 𝑥𝑥𝑛𝑛+1 = 𝑥𝑥𝑛𝑛 − for 𝑛𝑛 = 0, 1, 2, 3, 4, … 𝑓𝑓 𝑥𝑥𝑛𝑛 Downloaded from hlayiso.com www.futuremanagers.com

Module 3: Application of Differentiation (continued) MAXIMA AND MINIMA The maxima and minima (which are the plurals of maximum and minimum respectively) of a given function, which are collectively known as extrema, are the smallest and largest values of a function either in an interval or for the entire domain. Downloaded from hlayiso.com www.futuremanagers.com

Module 3: Application of Differentiation (continued) CONCAVITY AND INFLECTION POINT Concavity refers to the inward-curving or hollow object. Inflection point refers to a point of a curve at which a change in the direction of the curve occurs. Downloaded from hlayiso.com www.futuremanagers.com

Module 3: Application of Differentiation (continued) APPLICATION OF THE RATES OF CHANGE AND RELATED RATES Related rates problems can be solved by using the following procedure: • Define all the symbols you want to use in answering the question. • Write down all the relevant equations between the different variables. • Differentiate equations to reflect relationship to time and rates of change. • State the desired variable in terms of known variables. • Substitute values into the equation Downloaded and simplify. from hlayiso.com • Restate the answer in words to indicate the final result. www.futuremanagers.com

Module 4: Integration techniques INTRODUCTION Integration is used to find the area under the curve on a graph or to find the volume of an object Downloaded from hlayiso.com www.futuremanagers.com

Module 4: Integration techniques (continued) INTEGRATION BY INSPECTION 1. Guess the general form of the antiderivative. 2. Take the general form and differentiate it. 3. Compare the differentiated general form with the original integrand. 4. If the general form is correct but the answer is too large or small, put a multiplicative constant into the approximate form. 5. Add the constant of integration tofrom Downloaded the antiderivative. hlayiso.com 6. Differentiate your answer to verify that it gives you the original integrand. www.futuremanagers.com

Module 4: Integration techniques (continued) INTEGRATION BY MEANS OF ALGEBRAIC SUBSTITUTION 1. Find a substitution for 𝑢𝑢 = 𝑢𝑢(𝑥𝑥). 2. Find 𝑢𝑢 = 𝑢𝑢(𝑥𝑥) (the inverse of 𝑢𝑢(𝑥𝑥)). 3. Calculate the derivative: d𝑥𝑥 = 𝑥𝑥 ′ 𝑢𝑢 𝑑𝑑 .𝑑𝑑 4. Substitute the functions into the original integrand. 5. Simplify the function using normal algebraic techniques. 6. Check that theDownloaded new integral is easier from than the initial integral. hlayiso.com 7. Since 𝑓𝑓(𝑥𝑥)𝑑𝑑 is a function of 𝑥𝑥, substitute back to the initial variable 𝑥𝑥. 𝑑𝑑 www.futuremanagers.com

Module 4: Integration techniques (continued) INTEGRATION OF ALGEBRAIC FRACTIONS 1. Write down the given integral. 2. Come up with a substitution 𝑢𝑢 = 𝑢𝑢(𝑥𝑥). 3. Find the inverse function of 𝑢𝑢(𝑥𝑥), meaning that you will find 𝑢𝑢 = 𝑢𝑢(𝑥𝑥). 4. Differentiate to find 𝑑𝑑 = 𝑥𝑥 ′ 𝑢𝑢 𝑑𝑑 . 𝑑𝑑 𝑑𝑑 5. Do not forget that the answer to 𝑓𝑓(𝑥𝑥)𝑑𝑑 is a function of 𝑥𝑥. Therefore, once 𝑑𝑑 you have finished doing all your from Downloaded calculations, you should substitute back to hlayiso.com the initial variable 𝑥𝑥. www.futuremanagers.com

Module 5: Application of the definite integral INTRODUCTION When we solve an indefinite integral, the result is mostly another function; when we solve a definite integral, the result is often a numerical value. A definite integral always has a starting point and an end point between which the integral needs to be solved. Downloaded from hlayiso.com www.futuremanagers.com

Module 5: Application of the definite integral (continued) SOLVING THE DEFINITE INTEGRAL To determine a definite integral, we can integrate the function as before. Then we substitute the endpoint value into the independent variable of the integrate and subtract the integrate where the independent variable has been substituted with the start point value. Downloaded from hlayiso.com www.futuremanagers.com

Module 5: Application of the definite integral (continued) AREAS Whenever we want to calculate the area for a given interval of a function, we can use a definite integral. Downloaded from hlayiso.com www.futuremanagers.com

Module 5: Application of the definite integral (continued) VOLUMES OF SOLIDS AND REVOLUTION Volume is the amount of space that a substance or object occupies limited by boundaries on all its sides. Many solids have a circular cross-section in a plane perpendicular to some axis. These solids are referred to as solids of revolution because they are generated by rotating a region 360° about an axis. Downloaded from hlayiso.com www.futuremanagers.com

Module 5: Application of the definite integral (continued) SECOND MOMENT OF AREA The second moment of area also known as the moment of inertia of the plane area or the area moment of inertia is the geometric property of an area showing the points distributed randomly along a given axis. The second moment of area is determined with respect to an axis of rotation. Downloaded from hlayiso.com www.futuremanagers.com

Module 5: Application of the definite integral (continued) MASS MOMENT OF INERTIA (SECOND MOMENT OF MASS) The second moment of mass is an extension of the moment of inertia to the mass instead of area. The mass moment of inertia can loosely be defined as the property of a body that is central to reluctance of the body to move or accelerate. It can be denoted as: 2 Downloaded from𝑚𝑚hlayiso.com 𝐼𝐼 𝐴𝐴 = 𝑟𝑟 www.futuremanagers.com

Module 6: Differential equations INTRODUCTION Because derivatives are used to show rates of change, models are usually made up of a function and one or more of its derivatives. These equations 𝑑𝑑 are called differential equations (DE) and take the form = 𝑓𝑓 (𝑡𝑡) of the 𝑑𝑑 𝑑𝑑𝑑𝑑 𝑓 function 𝑓𝑓. Downloaded from hlayiso.com www.futuremanagers.com

Module 6: Differential equations (continued) PARTICULAR AND GENERAL SOLUTIONS OF DIFFERENTIAL EQUATIONS The differential equation y ′ = 𝑓𝑓(𝑥𝑥; 𝑦𝑦) takes a form which is simple if the function 𝑓𝑓 is independent of the dependent variable 𝑦𝑦: 𝑑𝑑𝑑𝑑 = 𝑓𝑓 𝑥𝑥 (equation 6.1) 𝑑𝑑𝑑𝑑 In this case, we need to integrate both sides of the equation to obtain, 𝑦𝑦 = ∫ 𝑓𝑓 𝑥𝑥 𝑑𝑑 + 𝐶𝐶 𝑑𝑑 Downloaded from hlayiso.com (equation 6.2) www.futuremanagers.com

Module 6: Differential equations (continued) SOLVE FIRST ORDER DIFFERENTIAL EQUATIONS There are two ways to find the general solution of the differential equation: 𝑑𝑑𝑑𝑑 1. Direct integration: = 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2. Separation of variables: = 𝑓𝑓 𝑥𝑥 𝑔𝑔(𝑦𝑦) 𝑑𝑑𝑑𝑑 Downloaded from hlayiso.com www.futuremanagers.com

Module 6: Differential equations (continued) SOLVE SECOND ORDER EQUATIONS Second order differential equations are differential equations that include a second derivative. They use the form: 𝑑𝑑 2 𝑦𝑦 𝑑𝑑 𝑄𝑄 𝑥𝑥 = 𝑎𝑎 2 + 𝑏𝑏 + 𝑐𝑐 𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑 𝑐𝑐 𝑑𝑑 Downloaded from hlayiso.com www.futuremanagers.com