Higher Mathematics Unit 1.3 Using Differentiation

Increasing & Decreasing Functions, Stationary Points A function is increasing when f ′ (x) > 0 A function is decreasing when f ′(x) < 0 Stationary points occur when f ′(x) = 0

Stationary Points The nature of a stationary point depends on the gradient on either side of it. Stationary points may be one of many types but always have f ′ (x) = 0 i.e. the gradient is 0 so the function is horizontal at that point.

Maximum Turning Point A maximum turning point (max TP) occurs when the gradient either side of a stationary point changes from positive to negative. T P x   f ′ (x) Graph Max TP + ve0− ve

Minimum Turning Point x  TP  f ′ (x) Graph A minimum turning point (min TP) occurs when the gradient either side of a stationary point changes from negative to positive. T P Min TP + ve0− ve

Rising Point of Inflexion x  TP  f ′ (x) Graph A rising point of inflexion (P I) occurs when the gradient either side of a stationary point remains positive. P I Rising P I + ve0

Falling Point of Inflexion x  TP  f ′ (x) Graph A falling point of inflexion (P I) occurs when the gradient either side of a stationary point remains negative. P I Falling P I + ve0− ve

Determine the Nature of Stationary Points Solve the equation f ′ (x) = 0 and consider the gradient f ′ (x) around each stationary point: If f ′ (x) is positive then the graph is increasing If f ′ (x) is negative then the graph is decreasing

Curve Sketching In order to sketch the graph of a function we need to find: the y-intercept the x-intercept the stationary points and their nature the behaviour of the curve for large positive and negative x

Curve Sketching It is important to justify your working for each stage of the process. The intercepts should be clearly marked on your final sketch. The coordinates of the turning points must be clearly labelled All relevant working must be noted and clearly understood. (0,0)(3,0) (2, -1)

Maximum & Minimum Values on a Closed Interval In a closed interval the maximum and minimum values of a function are either at a stationary point or at an end point of the interval. Local maximum Local minimum

Closed Interval A closed interval is often written as { −2 < x < 3} or [−2, 3] This interval would refer to the values of the function from −2 to 3 Maximum and minimum values are either at a stationary point or at an end point, so start by finding these points. Find f (−2) and f (3) Establish the stationary values for f ′ (x) in the interval from −2 to 3 and determine their nature. Find the value of the function at this point You are now in a position to state the local maximum or minimum value of the function within the given interval.

Problem Solving Differentiation can be used to solve problems which require maximum or minimum values. Problems typically cover topics such as areas, volumes and rates of change. They often involve having to establish a suitable formula in one variable and then differentiating to find a maximum or minimum value. This is known as Optimisation. It is important to check the validity of any solutions as often an answer is either nonexistent (e.g. a negative length or time) or outside an acceptable interval.

Optimisation Optimisation is term we use to describe a simple process of finding a maximum or minimum value for a given situation. Normally we would have to graph functions and then use our graph to establish a maximum or minimum. As we have learned previously, differentiation allows us to quickly find the required value and is the expected manner of solving problems like this in Higher mathematics. Drawing graphs would not be an acceptable solution.

Optimisation Problems posed will often involve more than one variable. The process of differentiation requires that we rewrite or re- arrange formulae so that there is only one variable – typically x. Questions are therefore multi-part where the first part would involve establishing a formula in x from a given situation, part 2 would involve the differentiation and validation of an acceptable answer, with part 3 the solution to the problem set. You would do well to note that although the first part is important, you can normally expect to complete the rest of the question even when you cannot justify a formula in part 1.

Speed and Acceleration Speed and acceleration are physical quantities which involve a rate of change. Because of this they can be found using differentiation. Speed is simply the rate of change of distance over time i.e. Speed = Dist / Time. If the distance is a known function, typically d(t), then speed is simply d ′ (t). This can be called v(t) where v is the speed in a straight line. Acceleration is the rate of change of speed over time i.e. Acceleration = change in Speed / Time. If the speed is a known function, typically d ′(t), a function of d(t), then the acceleration is the second derivative of the distance function d ′ ′(t) or more normally v ′ (t). For simplicity, travel is once again in a straight line. Problems involving these quantities often involve evaluating a derived function at a given time or finding a maximum speed from the derived function using stationary values.