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Okay, Ambe! Let's tackle this exam paper in English. QUESTION 1 a) Define the following terms: i) Function A function is a relation between a set of inputs (domain) and a set of permissible outputs (codomain), where each input is related to exactly one output. ii) Injective function An injective function (or one-to-one function) f: A B is a function where every element of the codomain is mapped to by at most one element of the domain. In other words, if f(x_1) = f(x_2), then x_1 = x_2. iii) Surjective function A surjective function (or onto function) f: A B is a function where every element of the codomain is mapped to by at least one element of the domain. In other words, for every y B, there exists at least one x A such that f(x) = y. b) State the following theorems: i) Maximum and Minimum Value Theorem (Extreme Value Theorem) If a function f is continuous on a closed and bounded interval [a, b], then f attains an absolute maximum value and an absolute minimum value on that interval. ii) Taylor Theorem If a function f is n times differentiable at a point a, then it can be approximated by a Taylor polynomial of degree n around a. Taylor's Theorem with Peano remainder states: f(x) = _k=0^n f^(k)(a)k!(x-a)^k + o((x-a)^n) iii) Lagrange Remainder Theorem The Lagrange Remainder Theorem provides a specific form for the remainder term in Taylor's Theorem. If f is a function that is (n+1) times continuously differentiable on an interval containing a and x, then the remainder R_n(x) = f(x) - _k=0^n f^(k)(a)k!(x-a)^k can be written as: R_n(x) = f^(n+1)(c)(n+1)!(x-a)^n+1 for some c between a and x. QUESTION 2 a) Define derivative of a function at a point and hence compute the derivative of a function f(x) = 3x^3 + 2x - 1 at x=2. The derivative of a function f at a point x_0 is the limit of the average rate of change of the function as the increment h approaches zero. It represents the slope of the tangent line to the graph of f at x_0. f'(x_0) = _h 0 (f(x_0+h) - f(x_0))/(h) To compute the derivative of f(x) = 3x^3 + 2x - 1 at x=2: Step 1: Find the general derivative of f(x). f'(x) = (d)/(dx)(3x^3 + 2x - 1) f'(x) = 3 · 3x^3-1 + 2 · 1x^1-1 - 0 f'(x) = 9x^2 + 2 Step 2: Evaluate the derivative at x=2. f'(2) = 9(2)^2 + 2 f'(2) = 9(4) + 2 f'(2) = 36 + 2 f'(2) = 38 b) Define composite function and hence if f: x 3x+1, g: x x-1, h: x 2x. Find (i) hgf (ii) fgh (iii) ghf and find the value of hgf(4), fgh(2) and ghf(-2). A composite function is a function formed by applying one function to the result of another function. If f and g are two functions, the composition f g (read "f of g") is defined by (f g)(x) = f(g(x)). Given functions: f(x) = 3x+1 g(x) = x-1 h(x) = 2x i) hgf(x) Step 1: Calculate gf(x). gf(x) = g(f(x)) = g(3x+1) gf(x) = (3x+1) - 1 gf(x) = 3x Step 2: Calculate hgf(x). hgf(x) = h(gf(x)) = h(3x) hgf(x) = 2(3x) hgf(x) = 6x ii) fgh(x) Step 1: Calculate gh(x). gh(x) = g(h(x)) = g(2x) gh(x) = (2x) - 1 gh(x) = 2x-1 Step 2: Calculate fgh(x). fgh(x) = f(gh(x)) = f(2x-1) fgh(x) = 3(2x-1) + 1 fgh(x) = 6x - 3 + 1 fgh(x) = 6x-2 iii) ghf(x) Step 1: Calculate hf(x). hf(x) = h(f(x)) = h(3x+1) hf(x) = 2(3x+1) hf(x) = 6x+2 Step 2: Calculate ghf(x). ghf(x) = g(hf(x)) = g(6x+2) ghf(x) = (6x+2) - 1 ghf(x) = 6x+1 Calculate the values: hgf(4) Using hgf(x) = 6x: hgf(4) = 6(4) = 24 fgh(2) Using fgh(x) = 6x-2: fgh(2) = 6(2) - 2 fgh(2) = 12 - 2 = 10 ghf(-2) Using ghf(x) = 6x+1: ghf(-2) = 6(-2) + 1 ghf(-2) = -12 + 1 = -11 c) Define inverse function and hence if f: x 2x+5, g: x x+4. Find (i) f^-1 (ii) g^-1 (iii) (fg)^-1 (iv) Show that g^-1f^-1 = (fg)^-1. An inverse function, denoted f^-1, is a function that "reverses" the action of another function f. If f(x)=y, then f^-1(y)=x. For a function to have an inverse, it must be bijective (both injective and surjective). Given functions: f(x) = 2x+5 g(x) = x+4 i) f^-1(x) Step 1: Set y = f(x). y = 2x+5 Step 2: Solve for x in terms of y. y-5 = 2x x = (y-5)/(2) Step 3: Replace y with x to get f^-1(x). f^-1(x) = (x-5)/(2) ii) g^-1(x) Step 1: Set y = g(x). y = x+4 Step 2: Solve for x in terms of y. x = y-4 Step 3: Replace y with x to get g^-1(x). g^-1(x) = x-4 iii) (fg)^-1(x) Step 1: Calculate fg(x). fg(x) = f(g(x)) = f(x+4) fg(x) = 2(x+4) + 5 fg(x) = 2x + 8 + 5 fg(x) = 2x+13 Step 2: Set y = fg(x). y = 2x+13 Step 3: Solve for x in terms of y. y-13 = 2x x = (y-13)/(2) Step 4: Replace y with x to get (fg)^-1(x). (fg)^-1(x) = (x-13)/(2) iv) Show that g^-1f^-1 = (fg)^-1 Step 1: Calculate g^-1f^-1(x). g^-1f^-1(x) = g^-1(f^-1(x)) We have f^-1(x) = (x-5)/(2) and g^-1(x) = x-4. g^-1f^-1(x) = g^-1((x-5)/(2)) g^-1f^-1(x) = ((x-5)/(2)) - 4 g^-1f^-1(x) = (x-5)/(2) - (8)/(2) g^-1f^-1(x) = (x-5-8)/(2) g^-1f^-1(x) = (x-13)/(2) Step 2: Compare with (fg)^-1(x). We found (fg)^-1(x) = (x-13)/(2). Since g^-1f^-1(x) = (x-13)/(2) and (fg)^-1(x) = (x-13)/(2), we can conclude that g^-1f^-1 = (fg)^-1. d) Define derivative of a function at a point and hence compute the derivative of a function f(x) = 3x^3 + 2x - 1 at x=2. This question is identical to question 2a. The solution is the same. The derivative of a function f at a point x_0 is the limit of the average rate of change of the function as the increment h approaches zero. It represents the slope of the tangent line to the graph of f at x_0. f'(x_0) = _h 0 (f(x_0+h) - f(x_0))/(h) To compute the derivative of f(x) = 3x^3 + 2x - 1 at x=2: Step 1: Find the general derivative of f(x). f'(x) = 9x^2 + 2 Step 2: Evaluate the derivative at x=2. f'(2) = 9(2)^2 + 2 = 36 + 2 = 38 QUESTION 3 a) Define integration of a function and hence solve the following integrals: Integration of a function is the process of finding the area under the curve of that function. It is the inverse operation of differentiation. The indefinite integral of a function f(x) is a function F(x) whose derivative is f(x), i.e., F'(x) = f(x). The definite integral calculates the net area between the function and the x-axis over a given interval. i. 2sqrt(x)\,dx Step 1: Rewrite sqrt(x) as x^1/2. 2x^1/2\,dx Step 2: Apply the power rule for integration x^n\,dx = x^n+1n+1 + C. 2 x^1/2\,dx = 2 · x^1/2+11/2+1 + C = 2 · x^3/23/2 + C = 2 · (2)/(3) x^3/2 + C = (4)/(3) x^3/2 + C ii. _3^8 (dx)/(sqrt(1+x)) Step 1: Rewrite the integral with a negative power. _3^8 (1+x)^-1/2\,dx Step 2: Perform a simple substitution. Let u = 1+x, then du = dx. Change the limits of integration: if x=3, u=1+3=4. If x=8, u=1+8=9. _4^9 u^-1/2\,du Step 3: Integrate using the power rule. [ u^-1/2+1-1/2+1 ]_4^9 = [ u^1/21/2 ]_4^9 = [ 2sqrt(u) ]_4^9 Step 4: Evaluate at the limits. 2sqrt(9) - 2sqrt(4) = 2(3) - 2(2) = 6 - 4 = 2 iii. (dx)/(2x^2+8x+20) Step 1: Factor the denominator to complete the square. 2x^2+8x+20 = 2(x^2+4x+10) = 2(x^2+4x+4+6) = 2((x+2)^2+6) The integral becomes: (dx)/(2((x+2)^2+6)) = (1)/(2) (dx)/((x+2)^2+6) Step 2: Perform a substitution. Let u = x+2, then du = dx. (1)/(2) (du)/(u^2+6) Step 3: Use the integration formula (1)/(a^2+x^2)\,dx = (1)/(a) ((x)/(a)) + C. Here a^2=6, so a=sqrt(6). (1)/(2) · (1)/(sqrt(6)) ((u)/(sqrt(6))) + C Step 4: Substitute back u = x+2. = (1)/(2sqrt(6)) ((x+2)/(sqrt(6))) + C b) State fundamental theorem of calculus and hence find the area bounded by the curves x=y^2 and x=2-y^2 on the closed interval [-1, 1]. The Fundamental Theorem of Calculus establishes a connection between the operations of differentiation and integration. It consists of two parts: Part 1: If f is continuous on [a, b], then the function F(x) = _a^x f(t)\,dt is continuous on [a, b] and differentiable on (a, b), and F'(x) = f(x) for all x (a, b). Part 2: If f is continuous on [a, b] and F is any antiderivative of f (i.e., F'(x) = f(x)), then _a^b f(x)\,dx = F(b) - F(a). To find the area bounded by the curves x=y^2 and x=2-y^2 on the closed interval [-1, 1]: We will integrate with respect to y. The interval [-1, 1] is for y. Step 1: Find the points of intersection of the two curves. y^2 = 2-y^2 2y^2 = 2 y^2 = 1 y = ± 1 The intersection points are y=-1 and y=1, which match the given interval boundaries. Step 2: Determine which function is "to the right" (has a larger x-value) on the interval [-1, 1]. Let's pick a test point y=0 within the interval: For x=y^2, x=0^2=0. For x=2-y^2, x=2-0^2=2. Since 2 > 0, the curve x=2-y^2 is to the right of x=y^2. Step 3: Set up the integral for the area. The area A is given by the integral of the right function minus the left function, integrated with respect to y. A = _-1^1 ((2-y^2) - y^2)\,dy A = _-1^1 (2 - 2y^2)\,dy Step 4: Integrate the function. A = [ 2y - (2y^3)/(3) ]_-1^1 Step 5: Evaluate the integral at the limits. A = ( 2(1) - (2(1)^3)/(3) ) - ( 2(-1) - (2(-1)^3)/(3) ) A = ( 2 - (2)/(3) ) - ( -2 - (2(-1))/(3) ) A = ( 2 - (2)/(3) ) - ( -2 + (2)/(3) ) A = 2 - (2)/(3) + 2 - (2)/(3) A = 4 - (4)/(3) A = (12)/(3) - (4)/(3) A = (8)/(3) That's 2 down. 3 left today — send the next one.