Question 4.c asks to find the values of k for which the function g(x) is continuous at x=2.

Okay, let's solve question 4.c. Question 4.c asks to find the values of k for which the function g(x) is continuous at x=2. The function is defined as: g(x) = k^2 x^2 & for x 2 \\ (1-k)x & for x > 2 For a function to be continuous at a point x=c, the following conditions must be met: 1. g(c) must be defined. 2. _x c^- g(x) must exist. 3. _x c^+ g(x) must exist. 4. _x c^- g(x) = _x c^+ g(x) = g(c). Here, c=2. Step 1: Determine g(2). For x 2, g(x) = k^2 x^2. Substituting x=2: g(2) = k^2 (2)^2 = 4k^2 The function is defined at x=2. Step 2: Evaluate the left-hand limit _x 2^- g(x). For x < 2, g(x) = k^2 x^2. _x 2^- g(x) = _x 2^- k^2 x^2 = k^2 (2)^2 = 4k^2 Step 3: Evaluate the right-hand limit _x 2^+ g(x). For x > 2, g(x) = (1-k)x. _x 2^+ g(x) = _x 2^+ (1-k)x = (1-k)(2) = 2 - 2k Step 4: Set the left-hand limit, right-hand limit, and g(2) equal for continuity. For g(x) to be continuous at x=2, we must have: _x 2^- g(x) = _x 2^+ g(x) = g(2) So, we set the expressions from Step 1, 2, and 3 equal to each other: 4k^2 = 2 - 2k Rearrange the equation into a standard quadratic form ax^2 + bx + c = 0: 4k^2 + 2k - 2 = 0 Divide the entire equation by 2 to simplify: 2k^2 + k - 1 = 0 Factor the quadratic equation: 2k^2 + 2k - k - 1 = 0 2k(k+1) - 1(k+1) = 0 (2k-1)(k+1) = 0 This gives two possible values for k: 2k-1 = 0 2k = 1 k = (1)/(2) k+1 = 0 k = -1 The values of k for which g(x) is continuous at x=2 are k = (1)/(2) and k = -1. 3 done, 2 left today. You're making progress.