Sets. 1.2 Find the set of all x R satisfying > = > = > = - > 0 = [x- 3 (x -2)] > 0. = - (x 1) (x 2) (x 3) > 0. Test x = 0, 5

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1 Sets 1.2 Find the set of all x R satisfying > > Test x 0, 5 > - > 0 [x- 3 (x -2)] > 0 - (x 1) (x 2) (x 3) > 0 At x0: y - (-1)(-2)(-3) 6 > 0 x < 1 At x5: y - (4)(3)(2) -24 < 0 2 < x < 3 Hence, {x R: x < 1 AND 2 < x < 3} 1.4 Find the domain of f(x), when f(x) Dom(f) { x R: x 2 4x + 3 0} { x R: (x-3)(x-1) 0} { x R: x 3 AND x 1} 1.4 Let f: R R and g: [0, ) R f(x) x 2-1 and g(x) x Show the composition of both, ensuring it remains a function. (f o g)(x) f(g(x)) 1 x -1 (g o f)(x) g(f(x)) Not defined for x0 hence must restrict domain Where: D { x R: x 2-1 0} { x R: x -1 AND x 1} Hence: g o f: D R is defined by: (g o f)(x)

2 Limits 2.1 Find, if it exists, Divide Top and Bottom by the highest power of x and take the lim: (cancel x 2 ) (Take limit, making all x-denominators approach infinite) 2.1 Using Pinching Theorem, find : Form an inequality of functions: -1 sin x 1 Now, for the (1/x): - Now, take the limit: 0 Hence, by pinching theorem, Find M, such that ε 0.5, for { Try to define M in terms of ε, using the definition: f(x) L <, where L 10: x-5 < M (as we are looking for the region (5-M, 5+M) To make this in the form above, we multiply by 2 to get the limit, L 10 (as 2x where x5 is 10) x-10 < 2M for x 5 Hence, 2M ε M, F(x) L < So, given the range that ε 0.5, M 0.5/2 Hence, the domain is , with the corresponding range being

3 Properties of Continuous Functions 3.2 Suppose that { where k is a real number. Find all values k such that f will be continuous everywhere. State: f is continuous for x 4 as is continuous. 8 for continuity at 4, k Show f x 3 5x + 3 has a zero in [0,1] f(0) 3 f(1) -1 Therefore, by IVT, there is some c in the interval [0,1] such that f(c) Does f(x)x 2-4 reach a max on [-3,5] State: f(x) is a polynomial and thus is continuous on a given closed interval; hence it attains a max in this interval. As the turning point is at (0,-4), it must reach a max at x5. Does f(x) x 2 4 reach a max on (-3,5) State: f(x) is a polynomial and thus is continuous; however the given interval is open. Hence the function does not reach a maximum within this interval. Differentiable Functions 4.1 Find the gradient of the tangent at x, for f(x) x 3 3x 2 + 3xh + h 3 3x 2

4 4.4 Find x 2 +2xy + y 2 100) 2x + (y 2 ) + 2 (xy) 0 2x + (2y. ) + 2(1.x + 1.y)( ) 2x +2y + (2x +2y)( So, 2x +2y (-2y-2x) Hence, Suppose f(x) Without a calculator, approximate f(8.01). Then find the equation of the tangent at (8,2), using this to find a better approximation. Basic approximate d/dx ( ) 1/3x 2 Eq of tangent at (8,2) then: 2 1/3x 2 (x-8) y1/3x - 8/3x 2 +2 Better approx. therefore is: via substitution into the equation. 4.7 A spherical balloon is being inflated and its radius is increasing at a constant rate of 6mm/sec. At what rate is its volume increasing when the radius of the balloon is 20mm?., where V4 r 3 /3, so dv/dt r 2 Hence, r 2 x 6 When r20, mm 3 /sec

5 Mean Value Theorem and Its Applications 5.1 Suppose that f: [1,8] R is given by f(x). Find a number c in (1,8) that satisfies the conclusions of the MVT for f on [1,8]. State: f is continuous on [1,8] and differentiable on (1,8). By the MVT there is a real number c in (1.8) such that: f (c) Now f (x) 1/3 x -2/3 so the above equation becomes; Rearranging for c: 7/3 So c (1,8), hence QED. 5.3 Use the MVT to show that ln(x) < x-1 whenever x > 1 State: Suppose that x > 1 and consider the interval [1,x]. We define a function f: [1,x] R by f(x) ln (t). Now f is continuous on [1,x] and differentiable on (1,x). So we may apply MVT. f'(t) 1/t, so: 1/c for some c in (1,x) by the MVT. That is, 1/c For some c between 1 and x. Since c > 1 we have 1/c < 1 and hence 1 Thus, ln (x) < x-1 QED.

6 5.4 Use MVT to find an upper bound for the error involved if we approximate by. State: The precise error is given by: -. We apply MVT to the function f, given by f(x), on the interval [25,26]. This gives: 1/2, for some c in (25, 26). Hence, Error 1/2 < 1/2 1/10, since c > 25. Hence, an upper bound for the error is 1/ Consider the parabolic arc: {(x,y) R 2 : yx 2 x 1, -2 x 3}. Find when the arc is closest to the origin. State: If (x,y) is a point on the arc, then its distance from the origin is given by. To find where the arc is closest, we want to minimise this equation, subject to the constraints yx 2 x 1 and -2 x 3. In other words, we want to find the global minimum for: f(x) in the interval [-2,3]. Let g(x) f(x 2 ), so we don t deal with square roots. Find the critical points of g(x): g'(x) 2(x-1)(x-1)(2x+1), hence critical points occur at -2, -1/2, 1 and 3. g(-2) 29; g(-1/2) 5/16; g(1) 2; g(3) 34. Hence, the arc is closest to the origin at x-1/ Determine the limiting behaviour of as x. State: Observe that both the numerator and denominator approach as x. Differentiating the numerator and denominator gives:, which again, as x, we have an indeterminate form of type. By differentiating again, we get:, and as x, f(x) 0. Hence, by L hopital s Rule:

7 Inverse Functions 6.1 Suppose g: (-,0] R is given by g(x) x Find g -1 (x) and define its domain and range. Find its derivative. g -1 (x) - Dom(g -1 ) [1, ] Range(g -1 ) (-,0] (g -1 ) (x) 6.4 Suppose that f: R R is given by f(x) x x+1. Find all possible intervals I of R, each as large as possible, such that the restricted function f: I R has an inverse. If x < 0 x -x If x > 1 x+1 -(x+1) If x < -1 f(x) (-x)(-(x+1)) x 2 + x If -1<x<0 f(x) (-x)(x+1) -x 2 x If x > 0 f(x) x(x+1) x 2 + x Hence, the inverse would exist in three intervals; (- -1], [-1,1], [1, ). 6.5 Prove the differential of cos -1 (x). ycos -1 (x) Hence, cos y x; where 0 < y < -sin (y) dy/dx 1 (by implicit differentiation) dy/dx 1/sin (y) cos (y) x hence cos 2 (y) x 2 sin 2 (y) + cos 2 (y) 1 sin 2 (y) 1 x 2 (where 0 < y <, sin(y) > 0) sin (y) Thus, d/dx (cos -1 (x)) QED.

8 Curve Sketching 7.1 Identify any vertical and oblique asymptotes of f(x). x -1, as denominator would be By polynomial long division of the function (x 2-2) divided by (x+1), we get x- Hence, there is a vertical asymptote at x0 and an oblique asymptote at yx Find the equation of the normal to the curve x y at the point P when t2. Using the x equation, x(t+1) t x + 1/t 1 (divide by t) 1/t 1-x Hence, t 1/1-x Thus, y 1/-x Differentiating, y (x) 1/x 2 Gradient of the Normal -1/m -x 2 Using Point-Gradient Formula y - ( ) - ( ) 2 (x - ) Sub in t2 Hence, the equation of the normal is: 27y -12x Convert (-1, -1) to Polar form with r > 0 and < θ <. r θ 3 rd quad, triangle sides 1, 1, x Pythagorean Triad 1, 1, (Right angle triangle) Angle to y axis - + Angle to ray 3 /4 Hence, (-1, -1) in polar form is (, -3 /4)

9 Integration 8.2 If f(x) x 2, find the upper and lower sums of f from 0 to 1. You may need to use the fact that: always given. if we need to use something like this is it State: We note that f is integrable on [0, 1] since it is bounded and continuous. Suppose that: Then P n {0, 1/n, 2/n, 3/n,.., n-1/n, n/n} (f) where f(x i ) f(a + i Δx) and Δx First, evaluate Δx using the limits b (1) and a (0), to find that Δx. Next, evaluate x i using the formula a + i Δx, to find x i. i Substitute this value for x i into f(x), giving (. i) 2 i 2. Put this into the upper sum formula: Multiply the 1/n terms, and take it out the front of the summation to get: Using the given summation of k 2, substitute in for i and take the limit: 1/3 Now taking the lower sum; x (i-1) (i-1. ) f(x i-1 ).

10 Substitute into the summation for k 2 ( ) 1/3 8.4 Find the area of the region bounded by the line yx and the parabola yx 2-2. Firstly, find any intersection points of the two curves: x x 2 2 x 2 x 2 0 (x 2)(x + 1) 0 Hence intersect at 2 and -1, making these our limits. This gives: [2-0.5] [-4/3 + 2/3] [1.5] [-2/3] 13/6 u Evaluate.dx Let u, hence du.dx Hence, 2 2e u + C 2 + C

11 8.9 Use integration by parts to evaluate Let x 7 dv and ln(x) U Hence, we integrate dv and differentiate U to get in the form: Thus, ln(x) - ln(x) Determine whether is convergent or divergent. By the limit definition, if does not exist, then the function diverges. Hence, take the definite integral of (2x-1). [x 2 -x] as x Hence the integral approaches making it divergent. The Logarithmic and Exponential Functions 9.5 Integrate tan x in respect to x -ln cos x +C 9.6 Differentiate 10 x ln (y) ln (10 x ) x ln (10) ln (10) by implicit differentiation Hence, y.ln(10) ln(10). 10 x

12 The Hyperbolic Functions 10.1 Sketch the curve defined by x(t) cosh t, y(t) sinh t. State: The parameter t can be eliminated by the hyperbolic identity: 1 cosh 2 x sinh 2 x (x(t)) 2 (y(t)) 2 x 2 y 2 1 This describes a parabola for x > 0. It has asymptotes yx and y -x 10.3 Prove the identity cosh(x+y) cosh x cosh y + sinh x sinh y State: By the definition of cosh, LHS ½ (e x+y + e -(x+y) ) RHS ½ (e x + e -x ) ½ (e y + e -y ) + ½ (e x - e -x ) ½ (e y - e -y ) ¼ (e x+y +e x-y +e -x+y +e -x-y ) + ¼ (e x+y - e x-y - e -x+y +e -x-y ) ½ (e x+y + e -(x+y) ) LHS (QED) 10.5 Simplify sinh(cosh -1 4/3) State: The hyperbolic identity can be rearranged to give sinh 2 t cosh 2 t 1. Substituting in the equation: Sinh 2 (cosh -1 4/3) (4/3) 2 1 7/9 Cosh > 0 hence we take the positive root: