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: $$ g(x) = \begin{cases} k^2 x^2 & \text{for } x \le 2 \\ (1-k)x & \text{for } x > 2 \end{cases} $$ For a function to be continuous at a point $x=c$, the following conditions must be met: 1. $g(c)$ must be defined. 2. $\lim_{x \to c^-} g(x)$ must exist. 3. $\lim_{x \to c^+} g(x)$ must exist. 4. $\lim_{x \to c^-} g(x) = \lim_{x \to c^+} g(x) = g(c)$. Here, $c=2$. Step 1: Determine $g(2)$. For $x \le 2$, $g(x) = k^2 x^2$. 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 $\lim_{x \to 2^-} g(x)$. For $x < 2$, $g(x) = k^2 x^2$. $$ \lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} k^2 x^2 = k^2 (2)^2 = 4k^2 $$ Step 3: Evaluate the right-hand limit $\lim_{x \to 2^+} g(x)$. For $x > 2$, $g(x) = (1-k)x$. $$ \lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (1-k)x = (1-k)(2) = 2 - 2k $$ 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: $$ \lim_{x \to 2^-} g(x) = \lim_{x \to 2^+} g(x) = g(2) $$ 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 $k$: $$ 2k-1 = 0 \implies 2k = 1 \implies k = \frac{1}{2} $$ $$ k+1 = 0 \implies k = -1 $$ The values of $k$ for which $g(x)$ is continuous at $x=2$ are $\boxed{k = \frac{1}{2} \text{ and } k = -1}$. 3 done, 2 left today. You're making progress.