Math Problem Statement
Solution
To determine the continuity and differentiability of the function at , we need to analyze the following conditions:
Step 1: Checking Continuity at
The function is defined as:
x, & \text{when } 0 \leq x < \frac{1}{2} \\ 1 - x, & \text{when } \frac{1}{2} \leq x < 1 \end{cases}$$ For $$f(x)$$ to be continuous at $$x = \frac{1}{2}$$, the following must hold: 1. The left-hand limit (LHL) as $$x \to \frac{1}{2}^-$$ should exist. 2. The right-hand limit (RHL) as $$x \to \frac{1}{2}^+$$ should exist. 3. The left-hand limit and right-hand limit should be equal to the value of the function at $$x = \frac{1}{2}$$. #### Calculate LHL Using $$f(x) = x$$ when $$x \to \frac{1}{2}^-$$: $$\lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^-} x = \frac{1}{2}$$ #### Calculate RHL Using $$f(x) = 1 - x$$ when $$x \to \frac{1}{2}^+$$: $$\lim_{x \to \frac{1}{2}^+} f(x) = \lim_{x \to \frac{1}{2}^+} (1 - x) = 1 - \frac{1}{2} = \frac{1}{2}$$ #### Evaluate $$f\left(\frac{1}{2}\right)$$ Since $$\frac{1}{2} \leq x < 1$$ applies here, use $$f(x) = 1 - x$$ to find: $$f\left(\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2}$$ Since $$\text{LHL} = \text{RHL} = f\left(\frac{1}{2}\right) = \frac{1}{2}$$, the function is continuous at $$x = \frac{1}{2}$$. ### Step 2: Checking Differentiability at $$x = \frac{1}{2}$$ For $$f(x)$$ to be differentiable at $$x = \frac{1}{2}$$, the left-hand derivative (LHD) and right-hand derivative (RHD) at $$x = \frac{1}{2}$$ must be equal. #### Calculate LHD $$\text{LHD} = \lim_{h \to 0^-} \frac{f\left(\frac{1}{2} + h\right) - f\left(\frac{1}{2}\right)}{h}$$ Using $$f(x) = x$$ as $$h \to 0^-$$: $$\text{LHD} = \lim_{h \to 0^-} \frac{\left(\frac{1}{2} + h\right) - \frac{1}{2}}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1$$ #### Calculate RHD $$\text{RHD} = \lim_{h \to 0^+} \frac{f\left(\frac{1}{2} + h\right) - f\left(\frac{1}{2}\right)}{h}$$ Using $$f(x) = 1 - x$$ as $$h \to 0^+$$: $$\text{RHD} = \lim_{h \to 0^+} \frac{\left(1 - \left(\frac{1}{2} + h\right)\right) - \frac{1}{2}}{h} = \lim_{h \to 0^+} \frac{\frac{1}{2} - h - \frac{1}{2}}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1$$ Since $$\text{LHD} = 1$$ and $$\text{RHD} = -1$$, the left-hand and right-hand derivatives are not equal. Thus, the function $$f(x)$$ is **not differentiable** at $$x = \frac{1}{2}$$. ### Conclusion - **Continuity**: $$f(x)$$ is continuous at $$x = \frac{1}{2}$$. - **Differentiability**: $$f(x)$$ is not differentiable at $$x = \frac{1}{2}$$. Would you like further details or have any questions? ### Related Questions 1. How does the continuity of a piecewise function differ from its differentiability? 2. What are the geometric interpretations of continuity and differentiability? 3. Can a function be continuous at a point but not differentiable at that point? 4. How would adding a constant to one piece of the function affect continuity? 5. What are some examples of functions that are continuous everywhere but not differentiable everywhere? ### Tip For piecewise functions, check continuity at boundary points by ensuring limits from both sides match the function's value.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Differentiability
Piecewise Function
Formulas
Left-hand limit (LHL): lim_{x → c^-} f(x)
Right-hand limit (RHL): lim_{x → c^+} f(x)
Left-hand derivative (LHD): lim_{h → 0^-} (f(c + h) - f(c)) / h
Right-hand derivative (RHD): lim_{h → 0^+} (f(c + h) - f(c)) / h
Theorems
Continuity and differentiability at a point
Suitable Grade Level
Grades 11-12
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