Show that the function \( f(x)=|x-6| \) is not differentiable at 6 . Find a formula for \( f^{\prime} \) and sketch its graph. The left-hand and right-hand derivatives of \( f \) at \( a \) are defined by \[ f_{-}^{\prime}(a)=\lim _{h \rightarrow 0^{-}} \frac{f(a+h)-f(a)}{h} \] and \( f_{+}^{\prime}(a)=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \) if these limits exist. Then \( f^{\prime}(a) \) exists if and only if these one-sided derivatives exist and are equal. (a) Find \( f_{-}^{\prime}(4) \) and \( f_{+}^{\prime}(4) \) for the function \[ f(x)=\left\{\begin{array}{ll}0 & \text { if } x \leqslant 0 \\ 5-x & \text { if } 0

Ask by Collins Bowman. in Pakistan

Dec 01,2024

To analyze the function \( f(x) = |x - 6| \), we first need to determine its differentiability at \( x = 6 \). ### Step 1: Check differentiability at \( x = 6 \) The function \( f(x) = |x - 6| \) can be expressed piecewise as: \[ f(x) = \begin{cases} 6 - x & \text{if } x < 6 \\ 0 & \text{if } x = 6 \\ x - 6 & \text{if } x > 6 \end{cases} \] To check if \( f \) is differentiable at \( x = 6 \), we need to compute the left-hand and right-hand derivatives. **Left-hand derivative at \( x = 6 \):** \[ f_{-}^{\prime}(6) = \lim_{h \to 0^{-}} \frac{f(6 + h) - f(6)}{h} = \lim_{h \to 0^{-}} \frac{(6 - (6 + h)) - 0}{h} = \lim_{h \to 0^{-}} \frac{-h}{h} = -1 \] **Right-hand derivative at \( x = 6 \):** \[ f_{+}^{\prime}(6) = \lim_{h \to 0^{+}} \frac{f(6 + h) - f(6)}{h} = \lim_{h \to 0^{+}} \frac{((6 + h) - 6) - 0}{h} = \lim_{h \to 0^{+}} \frac{h}{h} = 1 \] Since \( f_{-}^{\prime}(6) = -1 \) and \( f_{+}^{\prime}(6) = 1 \), the one-sided derivatives are not equal. Therefore, \( f \) is not differentiable at \( x = 6 \). ### Step 2: Find the derivative \( f' \) The derivative \( f' \) can be expressed as: \[ f'(x) = \begin{cases} -1 & \text{if } x < 6 \\ \text{undefined} & \text{if } x = 6 \\ 1 & \text{if } x > 6 \end{cases} \] ### Step 3: Sketch the graph of \( f \) The graph of \( f(x) = |x - 6| \) is a V-shaped graph with a vertex at \( (6, 0) \). The left side of the graph has a slope of -1, and the right side has a slope of 1. ### Step 4: Analyze the function \( f \) given in part (a) Now, let's analyze the function given in part (a): \[ f(x) = \begin{cases} 0 & \text{if } x \leq 0 \\ 5 - x & \text{if } 0 < x < 4 \\ \frac{1}{5 - x} & \text{if } x \geq 4 \end{cases} \] **(a) Find \( f_{-}^{\prime}(4) \) and \( f_{+}^{\prime}(4) \)** **Left-hand derivative at \( x = 4 \):** \[ f_{-}^{\prime}(4) = \lim_{h \to 0^{-}} \frac{f(4 + h) - f(4)}{h} = \lim_{h \to 0^{-}} \frac{(5 - (4 + h)) - (5 - 4)}{h} = \lim_{h \to 0^{-}} \frac{(1 - h) - 1}{h} = \lim_{h \to 0^{-}} \frac{-h}{h} = -1 \] **Right-hand derivative at \( x = 4 \):** \[ f_{+}^{\prime}(4) = \lim_{h \to 0^{+}} \frac{f(4 + h) - f(4)}{h} = \lim_{h \to 0^{+}} \frac{\frac{1}{5 - (4 + h)} - \frac{1}{1}}{h} = \lim_{h \to 0^{+}} \frac{\frac{1 - (5 - (4 + h))}{(5 - (4 + h))}}{h} = \lim_{h \to 0^{+}} \frac{h}{h(1)(1)} = \lim_{h \to 0^{+}} \frac{1}{1} = 1 \] Thus, \( f_{-}^{\prime}(4) = -1 \) and \( f_{+}^{\prime}(4) = 1 \). ### Step 5: Sketch the graph of \( f \) The graph of \( f \) consists of three segments: 1. A horizontal line at \( y = 0 \) for \( x \leq 0 \). 2. A line with a slope of -1 from \( (0, 5) \) to \( (4, 1) \). 3. A curve \( \frac{1}{5 - x} \) for \( x \geq 4 \), which approaches infinity as \( x \) approaches 5. ### Step 6: Analyze continuity and differentiability **(c) Where is \( f \) discontinuous?** The function is continuous everywhere except at \( x = 4 \) because the left-hand limit does not equal the right-hand limit at that point. **(d) Where is \( f \) not differentiable?** The function is not differentiable at \( x = 0 \) and \( x = 4 \) because the left-hand and right-hand derivatives do not match at these points. ### Summary - \( f(x) = |x - 6| \) is not differentiable at \( x = 6 \). - The derivative \( f'(x) \) is defined piecewise. - The function \( f \) in part (a) is discontinuous at \( x = 4 \) and not differentiable at \( x = 0 \) and \( x = 4 \).