QUESTION \( 4 \cdot 1 \) POINT Suppose that the]piecewise define function \( f \) is defined by \[ f(x)=\left\{\begin{array}{ll}1-2 x, & x \leq 3 \\ -2 x^{2}+4 x+1, & x>3\end{array}\right. \]

We are given the function   f(x) = { 1 – 2x  if x ≤ 3       –2x² + 4x + 1 if x > 3 } A common question with such a piecewise definition is to check whether f is continuous and differentiable at the point where the definition changes (here, at x = 3). Let’s work through these steps. 1. Check Continuity at x = 3   For f to be continuous at x = 3 we need:    limₓ→3⁻ f(x) = f(3) = limₓ→3⁺ f(x). a. Evaluate f(3):   Since f(x) = 1 – 2x when x ≤ 3, we have    f(3) = 1 – 2(3) = 1 – 6 = –5. b. Left-hand limit (x → 3⁻):   For x < 3, f(x) = 1 – 2x, so    limₓ→3⁻ f(x) = 1 – 2(3) = –5. c. Right-hand limit (x → 3⁺):   For x > 3, use f(x) = –2x² + 4x + 1, so    limₓ→3⁺ f(x) = –2(3)² + 4(3) + 1              = –18 + 12 + 1 = –5. Since all three values are equal (–5), f is continuous at x = 3. 2. Check Differentiability at x = 3   Even though f is continuous at x = 3, it might not be differentiable there. For differentiability at x = 3, the left and right derivatives must be equal. a. Compute the left derivative (x < 3):   For f(x) = 1 – 2x, the derivative is    f′(x) = –2.   Thus, the left-hand derivative at 3 is f′(3⁻) = –2. b. Compute the right derivative (x > 3):   For f(x) = –2x² + 4x + 1, differentiate:    f′(x) = –4x + 4.   Thus, at x = 3,    f′(3⁺) = –4(3) + 4 = –12 + 4 = –8. Since the left-hand derivative (–2) is not equal to the right-hand derivative (–8), f is not differentiable at x = 3. Summary of Answers: • f is continuous at x = 3. • f is not differentiable at x = 3 because the derivatives from the left and right are not equal.