Consider the following. \[ f(x)=\left\{\begin{array}{ll}x^{2}+3 & \text { if } x<1 \\ x+3 & \text { if } x \geq 1\end{array}\right. \] Is \( f \) differentiable at 1 ? Yes No Sketch the graph of \( f \).

To determine whether \( f \) is differentiable at \( x = 1 \), we first need to check if it is continuous at that point. The function is defined as \( f(x) = x^2 + 3 \) when \( x < 1 \) and \( f(x) = x + 3 \) when \( x \geq 1 \). Calculating the limit from the left as \( x \) approaches 1: \[ \lim_{x \to 1^-} f(x) = 1^2 + 3 = 4 \] Calculating the limit from the right as \( x \) approaches 1: \[ \lim_{x \to 1^+} f(x) = 1 + 3 = 4 \] Since both one-sided limits equal \( 4 \), \( f(1) = 4 \) ensures continuity at that point. Next, we examine differentiability by checking the derivatives from the left and right: \[ f'(x) = \begin{cases} 2x & \text{if } x < 1 \\ 1 & \text{if } x \geq 1 \end{cases} \] Calculating the left-hand derivative at \( x = 1 \): \[ \lim_{x \to 1^-} f'(x) = 2(1) = 2 \] Calculating the right-hand derivative at \( x = 1 \): \[ \lim_{x \to 1^+} f'(x) = 1 \] Since the left-hand derivative (2) does not equal the right-hand derivative (1), \( f \) is not differentiable at \( x = 1 \). No For the graph of \( f \), you'll see a parabola segment opening upwards starting at the point (1, 4) moving leftwards where the equation is \( f(x) = x^2 + 3 \). Right at that point, the graph transitions to a straight line with a slope of 1, starting at (1, 4) and moving upwards with the equation \( f(x) = x + 3 \). There will be a sharp corner at \( (1, 4) \), indicating it's continuous but not differentiable there.