Topic 2 Functions

Topic 2: Functions PART A Types of Relations: The above relation is a one-to-one relation as one input corresponds to one output. This is a one-to-many relation where one input corresponds to one or more outputs. 18 This is a many-to-one relation where many inputs correspond to the same specific output. There are also many-to-many relations, but they are not relevant to our subject. What is a function? A function is a mathematical method of relating one or more inputs to a specific output. For example: (Taking x as the input and y as the output) 𝑦 = 𝑥 2 is a function as one or two x-values correspond directly to one specific y-value, however, 𝑦 2 = 𝑥 is not a function as one x-value can correspond to one or more y-values. Essentially, only one-to-one relations and many-to-one relations are functions. There is a simple method of verifying if a particular graph is a function: The Vertical Line test. The graph above is given by𝑦 = 𝑥 2 and if we draw a vertical line through the graph at any xvalue, we will find that it corresponds to one specific y-value. For example, if we draw a line at x= -2, we get a y-value of 4 and if we draw a line at x=1, we get a y-value of 1. Thus, there is a specific output for every input. 19 Now, let us look at the graph of 𝑦 2 = 𝑥 Here, if we draw a vertical line at any x-value in the graph above, it will correspond to 2 outputs. The graph above illustrates a one-to-many relation. This is not a function as functions can only exist as one-to-one and many-to-one relations. Exercise: Is this graph a function? 20 1. 2. PART B Straight-Line Graphs: There are 3 different forms of straight-line graphs. 1. y = mx + c (gradient-intercept form). 2. ax + by + d = 0 (general form). 3. y − y1 = m(x − x1) (point-gradient form). 21 For two parallel lines: Gradient of Line 1= Gradient of Line 2 This can be simplified to be written as m1=m2 For two perpendicular lines: 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝐿𝑖𝑛𝑒 1 × 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝐿𝑖𝑛𝑒 2 = −1 PART C Domain and Range of Functions and Inverse Functions: The domain of a function consists of all the possible inputs/x-values that can make a function “work” and will result in the output of real y-values. For example, the function𝑦 = √3 − 𝑥 has the domain 𝑥 ≤ 3 as the square root of a negative number is not real. The range of a function consists of all the possible output values that result from the substitution of the domain into the function. For example, for the same function 𝑦 = √3 − 𝑥 , the yvalue/output is always greater than zero. Thus, the range can be written as 𝑦 ≥ 0. The inverse of a function is a function that undoes the action of another function. Graphically, this means that the inverse function is a reflection of the original function in the line y=x. In order to find the inverse of a function, we must swap y and x and then try to obtain y in terms of x again. For example, finding the inverse of 𝑦 = 𝑥 + 3. Swapping x and y: 𝑥 =𝑦+3 𝑦 =𝑥−3 Thus, 𝑦 = 𝑥 − 3 is the inverse of 𝑦 = 𝑥 + 3. This same principle can be applied to all functions. However, for the inverse of a function to be a function, the graph of the original function must pass the Horizontal Line Test Let us consider y = x2. 22 If we draw a horizontal line through any y-value of this graph, it will pass through 2 points. This is a problem as we eventually swap x and y while calculating the inverse of this function and that would give us a graph like the one shown below. As discussed earlier, this graph is not a function and thus, the inverse of y = x2 for all x-values is not a function. However, the inverse of y = x2 for 𝑥 ≥ 0is a function as all horizontal lines passing through the y-axis only pass through one output value. Thus, by restricting the domain of a function, we can ensure that the inverse of the function is a function as well. Exercise: 1. What is the domain and range of the function given below? 23 2. Find the inverse of this function and state and explain whether the inverse is a function or not. 𝑦 = (𝑥 − 1)4 − 4 PART D Quadratic Functions: A quadratic function can be expressed in 3 forms. 1. The standard form: y = ax2 + bx+ c 2. The vertex form: y = a(x-h)2+k where (h, k) is the vertex 3. The intercept form: y = a.(x-p).(x-q) where x=p and x=q where y=0, that is, p and q are the x-intercepts. Note- ‘a’ is the vertical stretch factor. Converting from one form to another: From standard form to vertex form: There are 2 possible ways of converting equations from standard form to vertex form. The first uses a method known as completing the square. Let us consider the equation y = x2 - 2x + 2. In this expression, a = 1. Thus, the method can be applied immediately. Steps: 24 1. Take the coefficient of the middle term, in this case, 2, divide it by 2 and square it. For this case, that gives us 1. 2. The answer from step 1 is added and subtracted on the same side of the equation. This is done because simply adding 1 on the same side of the expression will change its meaning, thus, to keep it the same, the same value must be subtracted on the same side 3. This gives us x2 - 2x + 2 +1 -1. Let us group together the perfect square expression, which is (x2 - 2x +1) and simplify the other two terms. Hence, we get y = (x2 - 2x +1) +1. 4. With further simplification, we obtain the equation in the vertex form: y= (x-1)2+1. The second method is much faster and can be used when you are more comfortable with quadratic equations. Let us consider the same equation y = x2 - 2x + 2. Here, a =1, b=-2 and c=2. To find the x-coordinate of the vertex, we can use the relation 𝑥 = −𝑏 2𝑎 .Then, we can obtain the y- value of the vertex by plugging in the x-value obtained from the above relation into the equation. Finally, we write the equation in the vertex form y = a(x-h)2+k using the values of ‘h’ and ‘k’ we just obtained. From vertex form to standard form: Simple expansion of the equation is required in order to convert the vertex form to the standard form. Considering the example of y= (x-3)2+2, we first square the term within the brackets. This gives us x2 -6x +9 +2 and that simplifies to x2 -6x +11. From standard form to intercept form: y = x -3x +2. The middle term is the sum of the two zeroes and the y-intercept ‘c’ is the product of the two zeroes. Thus, the two x-intercepts are x=1 and x=2 as they satisfy the two conditions mentioned earlier. 2 From intercept form to standard form: Simple expansion of the equation is required in order to convert the intercept form to the standard form. Considering the example of y= 2(x-1)(x-2), we first open multiply the terms within the brackets and then multiply all the terms in the simplified expression by 2. That will give a final equation as y= 2x2 -6x+4. From intercept form to vertex form: Let us take the equation: 𝑦 = (2𝑥 − 1)(𝑥 + 2). We will simplify it to :𝑦 = 2(𝑥 − 1/2)(𝑥 + 2). This means that x1 (the first x-intercept) =½ and x2 (the second x-intercept)= -2. Now, we shall add up x1 and x2. This gives us -3/2. This value is also equal to -b/a. Thus, to obtain the x-value for the vertex, we simply divide -3/2 by 2 and replace x by -¾ in our initial equation to get the yvalue of the vertex. Thus, we can rewrite the equation in the vertex form, using our calculations. From vertex form to intercept form: 25 There are two possible methods. We can reverse the methodology mentioned above and work backwards to obtain our equation in the intercept form. Alternatively, you can simply expand the vertex form to obtain the standard form and then use the standard form to intercept form conversion as mentioned earlier! Note: Sometimes, certain equations do not have x-intercepts and we will deal with them in depth later on. Domain and Range of Quadratic Equations: If we do not want to find the inverse of a quadratic function, its domain consists of the set of all real numbers. Contrarily, if we want to find the inverse function of a quadratic function, we must consider the x-coordinate of the vertex. If the inverse function is to be negative, then 𝑥 ≤ ℎ and if the inverse function is to be positive, then 𝑥 ≥ ℎ. The range of quadratic equations, however, depends on the sign of the ‘à’ value and the y-coordinate of the vertex. If 𝑎 > 0, the graph has a minimum value and thus, the range is 𝑦 ≥ 𝑘. However, if 𝑎 < 0, the graph has a maximum value and thus, the range is 𝑦 ≤ 𝑘. Using the Quadratic Formula: −𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 The above formula can be used with the standard form where y = ax2 + bx+ c. The part of this formula that is of particular interest to us is𝑏 2 − 4𝑎𝑐 𝑥= Let us understand the importance of this part, formally known as the discriminant. If 𝑏 2 − 4𝑎𝑐 >0, then a quadratic equation has 2 real, different roots (x-intercepts) If 𝑏 2 − 4𝑎𝑐 =0, then a quadratic equation has 2 real, equal roots (x-intercepts) If 𝑏 2 − 4𝑎𝑐<0, then a quadratic equation has no real roots and thus, has no x-intercepts. A very common question where utilising the discriminant is vital: 3𝑘𝑥 2 + 2𝑥 + 𝑘 = 0. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑘 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑔𝑖𝑣𝑒 2 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡, 𝑟𝑒𝑎𝑙 𝑟𝑜𝑜𝑡𝑠 𝑜𝑟 2 𝑒𝑞𝑢𝑎𝑙, 𝑟𝑒𝑎𝑙 𝑟𝑜𝑜𝑡𝑠 𝑜𝑟 𝑛𝑜𝑛𝑒, Here, we utilise the discriminant: 𝑏 2 − 4𝑎𝑐. We take ‘2’ as our ‘b’ value, ‘k’ as our c value and ‘3k’ as our ‘à’ value and use them to solve for the possible values of ‘k’ depending on the presence of real roots. Concavity: Most functions have a type of concavity. The concavity of quadratic functions depends on the sign of “a” in y = ax2 + bx+ c. If “a” is positive, then the function is positive and the graph of ‘f’ 26 is concave up. If “a” is negative, then the function is negative and the graph of ‘f’ is concave down. Exercise: Convert the following equations from the standard form to the vertex form, find its domain and range and simultaneously determine if the function has real roots: 1) 𝑦 = 𝑥 2 + 6𝑥 + 5 2) 𝑦 = 𝑥 2 + 3𝑥 + 2 3) 𝑦 = 3𝑥 2 + 12𝑥 + 19 Convert the following equations from the vertex form to the standard form, find its domain and range and simultaneously determine if the function has real roots: 1. 𝑦 = 3(𝑥 + 5)2 − 59/5 2. 𝑦 = 1/5 (𝑥 − 1)2 + 1 3. 𝑦 = 5/2 (𝑥 − 1/2)2 + 6 Convert the following equations from the standard form to the intercept form, find its domain and range and simultaneously determine if the function has real roots: 1. 𝑦 = 𝑥 2 + 13𝑥 + 36 2. 𝑦 = 2𝑥 2 + 8𝑥 + 6 3. 𝑦 = 4𝑥 2 − 9𝑥 + 2 Convert the following equations from the intercept form to the standard form, find its domain and range and simultaneously determine if the function has real roots: 1. 𝑦 = 2(𝑥 + 1)(x-5) 2. 𝑦 = (𝑥 + 1/9) (𝑥 − 3) 3. 𝑦 = 3(𝑥 + 2)(1 − 𝑥) Convert the following equations from the intercept form to the vertex form, find its domain and range and simultaneously determine if the function has real roots: 1. 𝑦 = 2(𝑥 + 1/2) (𝑥 − 3) 2. 𝑦 = (𝑥 − 1) (𝑥 + 3) 3. 𝑦 = 8(𝑥 + 1)(𝑥 − 2) Convert the following equations from the vertex form to the intercept form, find its domain and range and simultaneously determine if the function has real roots: 1. 𝑦 = (𝑥 + 2)2 − 1 2. 𝑦 = (𝑥 − 3)2 − 1 3. 𝑦 = 2(𝑥 − 1.25)2 − 6.125 27 PART E Graphing Functions and Using Graphs to obtain key data: Drawing a function: 2 Let us consider the function y=x . In order to draw the graph of y=x2, we must actually calculate the y-value of the function for every x-value, plot the points on a graph paper and draw the graph. Sketching a function: This is only an approximation of what the graph of a function will actually look like. For sketching a function, some key features that must be considered are the x-intercepts (roots), yintercepts, which can easily be calculated. While graphing functions, another key feature that must be considered is an asymptote. An asymptote is a value that the graph of a function keeps on getting closer and closer to, but never quite reaches. The most common example of asymptotes can be found in functions where ‘x’ is in the denominator. For example, let us 1 consider 𝑦 = 𝑥. At x=0, the value of y is infinite, which is not possible. Thus, the y-value approaches negative infinity as the x-value approaches zero from the left hand side and approaches positive infinity as the x-value approaches zero from the right hand side. This is perfectly illustrated in the graph below. 28 Other important aspects of a graph that must be considered are the maximum and minimum values. While some functions do not have maximum and minimum values, most do and they must be clearly illustrated in a graph. We have previously discussed maximum and minimum values through the vertex of a quadratic function. However, it is worth noting that the maximum and minimum values do not always have to determine the range of the function. Additionally, another key aspect we can find out from the graphs of functions, is their symmetry. Referring back to quadratic functions, their axis of symmetry lies along the x-value of their vertex as if the graph of the function is folded along that x-value, it will completely overlap. Exercise: Sketch the following graphs, keeping in mind the features previously discussed: 29 1. y=(x+3)(x-2)(x+5) 2. y= -(2x+1)2 1 3. 𝑦 = 𝑥−2 PART F Composite Functions: A composite function treats the output of one function as the input for another. This is represented by the notation (𝑓𝑜𝑔)(𝑥) or even f(g(x)). Let us consider an example. 𝑔(𝑥) = 𝑥 3 𝑓(𝑥) = 2𝑥 + 3 (𝑓𝑜𝑔)(𝑥) = 2𝑥 3 + 3 (𝑔𝑜𝑓)(𝑥) = (2𝑥 + 3)3 This clearly illustrates that (𝑓𝑜𝑔)(𝑥)is not necessarily equal to (𝑔𝑜𝑓)(𝑥). Additionally, composite functions only exist if the range of the first function is a subset of the domain of the second function. Another key aspect to note is that (f ∘ f -1)(x) = (f -1 ∘ f)(x) = x. This is a direct link to the content covered in Topic 2, Part B. Exercise: Find the composite functions, using f(x) and g(x) as noted below, and check if the composite function exists. 𝑓(𝑥) = √𝑥 + 3 2 𝑔(𝑥) = 5𝑥 + 6 1. (𝑓𝑜𝑓)(𝑥) 2. (𝑔𝑜𝑔)(𝑥) 3. (𝑓𝑜𝑔)(𝑥) 4. (𝑔𝑜𝑓)(𝑥) PART G Rational Functions: 𝑓(𝑥) A rational function is any function that can be expressed as a quotient 𝑦 = 𝑔(𝑥), g(x) is not equal to zero, where f(x) and g(x) are polynomial functions. Steps to graph rational functions: 1. Find the x-intercept by equating f(x) to zero 2. Find the y-intercept by substituting x=0 in both f(x) and g(x) 3. Find the vertical asymptote by equating g(x) to zero 4. Find the horizontal asymptote by dividing the coefficient of x in the numerator by the coefficient of x in the denominator 30 Some rational functions have oblique asymptotes. We can tell if a function has an oblique asymptote if we compare the highest power in the numerator to the highest power in the denominator. If the highest power in the numerator>highest power in the denominator, then the function has an oblique asymptote. The Math AA syllabus only includes functions of the type .To find the oblique asymptote, we divide the numerator by the denominator using long division.The quotient is the oblique asymptote. Additionally, this type of rational functions do not have horizontal asymptotes. However, rational functions of the type have their horizontal asymptote at y=0. This can be generalised for all functions where the degree of the numerator of f(x) is less than the degree of the denominator. However, if there is a vertical transformation ‘b’(mentioned in Part J), then the horizontal asymptote is y=b. Exercise 2G: Sketch the rational functions listed below, using the steps outlined above. 1. 𝑦 = 2. 𝑦 = 3. 𝑦 = 1 2𝑥−3 𝑥+4 𝑥−2 2𝑥+3 𝑥−1 PART H Exponential and Logarithmic Functions: Exponential Functions are expressed in the form: f(x) = ax , a > 0, or ex. Let us graph 2x. There is no x-intercept and there is a y-intercept at (0,1). The y-value doubles every time the x-value increases by 1. The graph obtained is given below: 31 All exponential functions have a y-intercept at (0,1) and will approach 0 as the x-value gets infinitely more negative as seen above. There is an exception to this rule, which will be mentioned later on. Logarithmic Functions are expressed in the form f(x) = logax, x > 0 or f(x) = lnx, x > 0. Let us graph log2x. There is a x-intercept at (1,0) and there is no y-intercept. Every time the xvalue doubles, the y-value increases by 1.(exactly the opposite of 2x, or mathematically, its inverse). The graph obtained is given below. 32 All logarithmic functions have a x-intercept at (1,0) and will approach negative infinity as the xvalue approaches 0. Exercise 2H: Graph the following exponential and logarithmic functions: 1. 𝑦 = 𝑙𝑛(𝑥) 2. 𝑦 = 4𝑥 3. 𝑦 = 𝑙𝑜𝑔(𝑥) 4. 𝑦 = 1𝑥 PART I Solving equations: Let us try and solve the equation given below. 22𝑥 − 4. 2𝑥 + 4 = 0 There are 2 methods that can be used. Firstly, we can simply graph this function on a calculator or a graphing software like Desmos and obtain the answer by checking the roots of the function. Alternatively, we can equate 2x to b and rewrite the equation as a quadratic equation, which is noted below: 𝑏 2 − 4𝑏 + 4 = 0 Once we have this equation, we can easily solve it to get the answer. Then, we can resubtitute that value of b in its relation to 2x and get the value of x. 33 However, there are some equations that we cannot solve algebraically and thus, must use technology to find their solution. One such equation is 𝑒 𝑥 = 𝑐𝑜𝑠(𝑥). Here, we simply graph ex and cos(x) and find their intersection on a calculator or graphing software. Exercise: Solve the following equations, either algebraically or graphically: 1. 𝑒 −𝑥 + 𝑒 𝑥 = 4 2. 𝑙𝑛(𝑥) = 𝑡𝑎𝑛(𝑥) PART J Basic Transformations of Functions: Transformations of functions are used to shift and/or resize the graphs of functions. There are 3 types of transformations that we will be looking at in depth. 1. Translations: y = f(x) + b; y = f(x − a). The first translation mentioned, f(x)+b essentially moves the graph of a function up or down along the y-axis. For example, if f(x)=x2 and b = -1, then the function x2 is moved down and has a y-intercept of -1 and there is a change in the xintercepts as well. The second translation has the same principle, but it moves the graph of a function left or right along the x-axis. So if a>0, the graph is translated to the right by ‘à’ units and if a<0, the graph is translated to the left by ‘à’ units. Let us consider an example. 𝑓(𝑥) = 𝑥 3 .The initial graph looks like this: However, if we want to graph the function 𝑦 = 𝑓(𝑥 − 1) + 1, the graph we obtain is shifted one unit to the right along the x-axis and moved up by one unit along the y-axis. This is what the resultant graph looks like: 34 We can apply the same principle to all functions. 2. Reflections in both axes: 𝑦 = − 𝑓(𝑥); 𝑦 = 𝑓( − 𝑥). 𝑦 = − 𝑓(𝑥)reflects the function f(x) over the x-axis. You essentially have to fold the graph over the x-axis. Let us consider the same example as mentioned earlier 𝑦 = (𝑥 − 1)3 + 1. The reflection of the graph in the x-axis is shown alongside the original graph. 35 Now, let us look at 𝑦 = 𝑓( − 𝑥). 𝑦 = 𝑓( − 𝑥)reflects the function f(x) over the y-axis. You essentially have to fold the graph over the y-axis. Using the same example, the reflection of the graph in the y-axis is shown alongside the original graph. However, there are certain functions where f(x)=f(-x). These types of functions are known as even functions. A simple example of this is 𝑓(𝑥) = 𝑥 2 . In this function, if we substitute ‘x’ with ‘-x’, we will get the same output function. Odd functions, on the other hand, are functions where f(-x) = f(x). An example for this is 𝑓(𝑥) = 𝑥 3 . If we substitute ‘-x’ in ‘x’, we get 𝑓(−𝑥) = −𝑥 3 which is equal to -f(x). 3. Dilations of functions: In the previous transformations, we only adjusted the ‘position’ of the graph of the function. In this type of transformation, we will actually change the shape of the graphs. A vertical stretch with scale factor ‘p’ is noted as y = pf(x). When p>1, the graph is vertically stretched, that is, the graph is pulled away from the x-axis. We find the coordinates of the graph by multiplying each of its y-coordinates by p. If 01, the graph is horizontally stretched, that is, the graph is pulled away from the y-axis. We find the coordinates of the graph by dividing each of its y-coordinates by q. If 0 g(x) between m and p, then g(x) is greater than f(x) for x>p and x0 And when a<0, the graph of 𝑦 = 𝑎𝑥 3 has the shape given below. 44 As a increases, the graph of y becomes steeper. Using our learnings from transformations, we recognise that the graph of 𝑦 = 𝑎(𝑥 − 𝑏)3 + 𝑐 is produced by translating the original graph of y by b units to the right/left and c units upwards/downwards. All cubic polynomials have at least one real zero and thus, must cut the x-axis at least once. Here is some additional information about the zeroes of cubic polynomials which will be immensely helpful in graphing them. For a cubic of the form 𝑃(𝑥) = 𝑎(𝑥 − 𝛼)(𝑥 − 𝛽)(𝑥 − 𝛾)where 𝛼, 𝛽, 𝛾 ∈ 𝑅, the graph has three distinct x-intercepts corresponding to the three distinct roots 𝛼, 𝛽, 𝑎𝑛𝑑 𝛾. For a cubic of the form 𝑃(𝑥) = 𝑎(𝑥 − 𝛼)2 (𝑥 − 𝛽)where 𝛼, 𝛽 ∈ 𝑅,the graph touches the x-axis at 𝛼 and 𝛽. For a cubic of the form 𝑃(𝑥) = 𝑎(𝑥 − 𝛼)3 , 𝑥 ∈ 𝑅,the graph has only one x-intercept, 𝛼. The graph is horizontal at this point. 45 For a cubic in the form 𝑃(𝑥) = 𝑎(𝑥 − 𝛼)(𝑎𝑥 2 + 𝑏𝑥 + 𝑐)where the discriminant is negative, there is only one x-intercept, 𝛼. For quartic polynomials, 𝑦 = 𝑎𝑥 4 .If a > 0, the graph opens upwards and if a<0, the graph opens downwards. If a quartic is factored into real linear factors then: 46 1. 2. 3. 4. 5. 6. Exercise: Find all real quartic polynomials with zeros: a. -1, 2, 1 ± √5 b. 3 ± 𝑖, ±√13 c. ±𝑖√5, 2, 9 Show that 𝑥 4 + 16 is divisible by two real quadratic factors that have the form 𝑥 2 + 𝑎𝑥 + 4 and 𝑥 2 + 𝑏𝑥 + 4, but cannot be divided into two real quadratic factors that have the form 𝑥 2 + 𝑎𝑥 + 8and 𝑥 2 + 𝑏𝑥 + 2. When P(x) is divided by x2-2x-3, the quotient is x2+3x+2 and the remainder R(x) is unknown. When P(x) is divided by x-3, the remainder is 12. When P(x) is divided by x-2, the remainder is 80. Find R(x) in the form ax+b 3+i is a root of 𝑥 4 + 𝑎𝑥 3 − 6𝑥 2 − 4𝑎𝑥 − 𝑏 = 0. Find a and b and deduce the other roots of the equation. A real quartic polynomial has a leading coefficient of 3 and constant term 20. Its zeros exist in the form 2𝑎 ± 𝑖 and 5 ± 𝑎 where a is a real number. Find the possible values of a. Find the equation of the quartic whose graph: 1 a. Touches the x-axis at 2 and cuts the x-axis at -3 and 4and passes through the point (-2, -14) 2 1 b. Cuts the x-axis at ± 5and 3and passes through the point (-3, 5) c. Passes through (-4, 14), (9,5), (-2,1) and (4,3) IB Questions: 1. (Non-Calculator) The function g(x) is defined by g(x) = x3 - 3x2 + 8x - 24 where x∈R (a) Find the remainder when g(x) is divided by (i) (x - 2) (ii) (x - 3). (b) Prove that g(x) has only one real zero. (c) Write down the transformation that will transform the graph of y = g(x) onto the graph of y = 8x3 - 12x2 + 16x - 24. 2. (Non-Calculator) Consider the following two functions: 2𝑥 2 + 3 𝑏(𝑥) = , 𝑥 ≥ 0. 75 |3𝑥 − 4| 𝑐(𝑥) = , 𝑥𝜖𝑅 10 (a) State the range of b(x) and c(x) and graph the two functions. (b) Find an expression for the composite function (𝑏𝑜𝑐)(𝑥) in the form 𝑝𝑥 2 +𝑞𝑥+𝑟 3750 where p , 47 q and r are integers. (c) (i) Find an expression for the inverse term 𝑓 −1 (𝑥). (ii) State the domain and range of 𝑓 −1 (𝑥) and graph the function. 3. (With Calculator) Find the set of values of x for which |0.1x2-2x+3|=log10(x) 4. The functions p and q are defined by 𝑝(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐and 𝑞(𝑥) = 𝑑𝑠𝑖𝑛(𝑥) + 𝑒𝑥 + 𝑓, 𝑥𝜖𝑅 where a, b, c, d, e, and f are real constants. a. Given that p(x) is an even function, show that b=0. b. Given that q(x) is an even function, find the value of f. c. The function r(x) is both even and odd, with domain R. Find h(x). 𝑥+𝑎 5. The function g(x) is of the form 𝑔(𝑥) = 𝑏𝑥+𝑐 , 𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 −𝑐 𝑏 .given that the graph 2 of f has asymptotes x = −4 and y = −2 , and that the point (3,1) lies on the graph, find the values of a , b and c . 6. (Non-Calculator) 7 a. Sketch on the same axes the curve 𝑦 = |𝑥−4|and the line y=x+2, clearly indicating any axes intercepts and any asymptotes. 7 b. Find the exact solutions to the equation 𝑥 + 2 = |𝑥−4| 7. The quadratic equation 𝑥 2 − 2𝑘𝑥 + (𝑘 − 1) = 0 has roots 𝛼 𝑎𝑛𝑑 𝛽 such that 𝛼2 + 𝛽 2 = 4. Without solving the equation, find the possible values of the real number k. 8. In the quadratic equation 7𝑥 2 − 8𝑥 + 𝑝 = 0, 𝑝𝜖𝑄, one root is three times the other. Find the value of p. 9. Let r(x)= ln(x). The graph of r(x) is transformed into the graph of the function s by a translation of 3 units to the right and 2 units downwards, followed by a reflection in the xaxis. Find an expression for s(x), giving your answer as a single logarithm. 10. Find the values of k such that the equation 𝑥 3 + 𝑥 2 − 3𝑥 + 1 = 𝑘 has 3 distinct real solutions. 11. (With Calculator)The graph of 𝑦 = 𝑙𝑛(𝑥)is transformed into the graph of 𝑦 = 𝑙𝑛(2𝑥 + 1). a. Describe the two transformations that are required to do so b. Solve 𝑙𝑛(2𝑥 + 1) > 3𝑠𝑖𝑛(𝑥), 𝑥𝜖[0,10]. 1 12. Consider the function 𝑓(𝑥) = 𝑥 2 +3𝑥+2 , 𝑥𝜖𝑅, 𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 − 2 𝑜𝑟 − 1. a. Express x2+3x+2 in the form (x+h)2+k b. Factorise x2+3x+2 c. Sketch the graph of f(x), indicating on it the equations of the asymptotes, the coordinates of the y-intercept and the local maximum. 1 1 1 d. Show that 𝑥 2 +3𝑥+2 = 𝑥+1 − 𝑥+2 e. Sketch the graph of f(|x|). 48