Sketch the graph of I and celermine where \( f \) is nondifforentable. \[ f(x)=\left\{\begin{aligned} -4 x & \text { if } x<1 \\ -4 & \text { if } x \geq 1 \end{aligned}\right. \] Where is the function \( f(x) \) nondifferuitable? Select the correct choice below and, if necessary, fill in the answer box to cnmplelo your choice. A. The funclion \( I(x) \) is nondifferentiable ai \( x= \) \( \square \) because the function is not continuous thero. B. The function \( l(x) \) is differentiable lor all real numbers C. The function \( f(x) \) is nondifferentiable at \( x=\square \) because there is a sharp turn there. D. The function \( f(x) \) is nondifferonliablo ot \( x=\square \) \( \square \) because there is a vertical langent line there.

Let's analyze the function \( f(x) \): \[ f(x) = \begin{cases} -4x & \text{if } x < 1 \\ -4 & \text{if } x \geq 1 \end{cases} \] ### **Graph of \( f(x) \):** 1. **For \( x < 1 \):** The function is a straight line with a slope of \(-4\), passing through the origin. 2. **For \( x \geq 1 \):** The function is a constant value \(-4\), which is a horizontal line. At \( x = 1 \): - **Left-hand limit:** \( f(1^-) = -4(1) = -4 \) - **Right-hand limit:** \( f(1^+) = -4 \) Since both limits are equal and equal to \( f(1) \), the function is **continuous** at \( x = 1 \). ### **Differentiability:** - **For \( x < 1 \):** The derivative of \( f(x) = -4x \) is \( f'(x) = -4 \). - **For \( x \geq 1 \):** The derivative of \( f(x) = -4 \) is \( f'(x) = 0 \). At \( x = 1 \): - **Left-hand derivative:** \( f'(1^-) = -4 \) - **Right-hand derivative:** \( f'(1^+) = 0 \) Since the left-hand derivative does not equal the right-hand derivative, \( f(x) \) is **not differentiable** at \( x = 1 \). ### **Conclusion:** The function \( f(x) \) has a sharp turn at \( x = 1 \), making it nondifferentiable at that point. **Correct Choice:** **C.** The function \( f(x) \) is nondifferentiable at \( x = 1 \) because there is a sharp turn there.