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Given the function y = f(x) = (x)/((x-1)(x-2)). i) The domain of definition of f. Step 1: Identify restrictions for the function. The function is a rational function, so the denominator cannot be zero. (x-1)(x-2) 0 Step 2: Solve for the values of x that make the denominator zero. x-1 = 0 x = 1 x-2 = 0 x = 2 Step 3: State the domain. The domain of f is all real numbers except x=1 and x=2. D_f = R \1, 2\ = (-, 1) (1, 2) (2, +) The domain of definition of f is R \1, 2\. ii) The limits at the bounds of the domain. Step 1: Calculate the limits as x ± . _x ± f(x) = _x ± (x)/((x-1)(x-2)) = _x ± (x)/(x^2 - 3x + 2) Divide the numerator and denominator by the highest power of x in the denominator, which is x^2: _x ± (1)/(x)1 - (3)/(x) + (2)/(x^2) = (0)/(1 - 0 + 0) = 0 So, _x - f(x) = 0 and _x + f(x) = 0. Step 2: Calculate the limits as x 1. As x 1, the numerator approaches 1, and the denominator approaches 0. This indicates a vertical asymptote. For x 1^- (e.g., x=0.9): Numerator x 1 (positive). Denominator (x-1) 0^- (negative). Denominator (x-2) -1 (negative). So, _x 1^- f(x) = (1)/((0^-)(-1)) = (1)/(0^+) = +. For x 1^+ (e.g., x=1.1): Numerator x 1 (positive). Denominator (x-1) 0^+ (positive). Denominator (x-2) -1 (negative). So, _x 1^+ f(x) = (1)/((0^+)(-1)) = (1)/(0^-) = -. Step 3: Calculate the limits as x 2. As x 2, the numerator approaches 2, and the denominator approaches 0. This indicates a vertical asymptote. For x 2^- (e.g., x=1.9): Numerator x 2 (positive). Denominator (x-1) 1 (positive). Denominator (x-2) 0^- (negative). So, _x 2^- f(x) = (2)/((1)(0^-)) = (2)/(0^-) = -. For x 2^+ (e.g., x=2.1): Numerator x 2 (positive). Denominator (x-1) 1 (positive). Denominator (x-2) 0^+ (positive). So, _x 2^+ f(x) = (2)/((1)(0^+)) = (2)/(0^+) = +. iii) The asymptotes, stationary points, the variation table and sketch the graph of the function y=f(x). Step 1: Determine the asymptotes. From the limits: • Vertical asymptotes: x=1 and x=2. • Horizontal asymptote: y=0 (as x ± ). There are no oblique asymptotes. Step 2: Find the stationary points. First, find the derivative f'(x) using the quotient rule ((u)/(v))' = (u'v - uv')/(v^2). Let u = x u' = 1. Let v = (x-1)(x-2) = x^2 - 3x + 2 v' = 2x - 3. f'(x) = (1 · (x^2 - 3x + 2) - x · (2x - 3))/((x^2 - 3x + 2)^2) f'(x) = (x^2 - 3x + 2 - 2x^2 + 3x)/((x^2 - 3x + 2)^2) f'(x) = (-x^2 + 2)/((x^2 - 3x + 2)^2) Set f'(x) = 0 to find critical points: -x^2 + 2 = 0 x^2 = 2 x = ± sqrt(2) Now, find the corresponding y-values: For x = sqrt(2): f(sqrt(2)) = sqrt(2)(sqrt(2)-1)(sqrt(2)-2) = sqrt(2)2 - 2sqrt(2) - sqrt(2) + 2 = sqrt(2)4 - 3sqrt(2) Rationalize the denominator: f(sqrt(2)) = sqrt(2)(4 + 3sqrt(2))(4 - 3sqrt(2))(4 + 3sqrt(2)) = 4sqrt(2) + 3(2)16 - 9(2) = 4sqrt(2) + 616 - 18 = 4sqrt(2) + 6-2 = -2sqrt(2) - 3 So, one stationary point is (sqrt(2), -2sqrt(2)-3) ≈ (1.41, -5.83). For x = -sqrt(2): f(-sqrt(2)) = -sqrt(2)(-sqrt(2)-1)(-sqrt(2)-2) = -sqrt(2)(sqrt(2)+1)(sqrt(2)+2) = -sqrt(2)2 + 2sqrt(2) + sqrt(2) + 2 = -sqrt(2)4 + 3sqrt(2) Rationalize the denominator: f(-sqrt(2)) = -sqrt(2)(4 - 3sqrt(2))(4 + 3sqrt(2))(4 - 3sqrt(2)) = -4sqrt(2) + 3(2)16 - 9(2) = -4sqrt(2) + 616 - 18 = -4sqrt(2) + 6-2 = 2sqrt(2) - 3 So, the other stationary point is (-sqrt(2), 2sqrt(2)-3) ≈ (-1.41, -0.17). Step 3: Construct the variation table. The sign of f'(x) = (-x^2 + 2)/((x^2 - 3x + 2)^2) is determined by the numerator -x^2 + 2, as the denominator is always positive for x 1, 2. -x^2 + 2 > 0 x^2 < 2 -sqrt(2) < x < sqrt(2). -x^2 + 2 < 0 x^2 > 2 x < -sqrt(2) or x > sqrt(2). The critical points are -sqrt(2) ≈ -1.41, 1, sqrt(2) ≈ 1.41, 2. | x | - | | -sqrt(2) | | 1 | | sqrt(2) | | 2 | | + | | :------------- | :-------- | :----------- | :--------------- | :----------- | :----------- | :----------- | :--------------- | :----------- | :----------- | :----------- | :-------- | | Sign of -x^2+2 | - | - | 0 | + | + | + | 0 | - | - | - | - | | Sign of f'(x) | - | - | 0 | + | Undefined | + | 0 | - | Undefined | - | - | | Variation of f(x) | 0 | | 2sqrt(2)-3 (Min) | | - | | -2sqrt(2)-3 (Max) | | + | | 0 | | | | | ≈ -0.17 | | + | | ≈ -5.83 | | - | | | Step 4: Sketch the graph. Based on the analysis: • The graph approaches the horizontal asymptote y=0 from above as x -. • It decreases to a local minimum at (-sqrt(2), 2sqrt(2)-3) ≈ (-1.41, -0.17). • It then increases, crossing the x-axis at x=0 (since f(0)=0), and approaches + as x 1^-. • Between x=1 and x=2, the graph starts from - as x 1^+. • It increases to a local maximum at (sqrt(2), -2sqrt(2)-3) ≈ (1.41, -5.83). • It then decreases and approaches - as x 2^-. • For x > 2, the graph starts from + as x 2^+. • It decreases and approaches the horizontal asymptote y=0 from above as x +. (A visual sketch cannot be provided in this text format, but the description above outlines the shape and key points for plotting.) What's next?