Math Problem Statement
Show that the function f(x) = { x/(1+x) if x ≥ 0, x^2 if x < 0 } is not differentiable at x = 0. Use the limit definition.
Solution
The problem asks to show that the given piecewise function:
\frac{x}{1+x} & \text{if } x \geq 0, \\ x^2 & \text{if } x < 0, \end{cases}$$ is not differentiable at $$x = 0$$ using the limit definition of the derivative. ### Step-by-Step Solution: The derivative of $$f(x)$$ at $$x = 0$$ is given by the limit definition of the derivative: $$f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h}.$$ First, let's evaluate $$f(0)$$ using the piecewise definition. - For $$x \geq 0$$, $$f(x) = \frac{x}{1+x}$$. So, $$f(0) = \frac{0}{1+0} = 0$$. - For $$x < 0$$, $$f(x) = x^2$$. At $$x = 0$$, this also gives $$f(0) = 0^2 = 0$$. Thus, $$f(0) = 0$$. Now, compute the limit for $$h \to 0^+$$ (approaching 0 from the right): $$f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h \to 0^+} \frac{h}{h(1+h)} = \lim_{h \to 0^+} \frac{1}{1+h} = 1.$$ Next, compute the limit for $$h \to 0^-$$ (approaching 0 from the left): $$f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h^2 - 0}{h} = \lim_{h \to 0^-} h = 0.$$ Since $$f'(0^+) = 1$$ and $$f'(0^-) = 0$$, the left-hand and right-hand limits are not equal. Therefore, $$f'(0)$$ does not exist, meaning the function is **not differentiable at $$x = 0$$**. Would you like further details or explanations on any step? Here are some questions to consider: 1. What is the general concept behind the limit definition of a derivative? 2. Why do we need to evaluate limits from both sides to check differentiability at $$x = 0$$? 3. How does the piecewise nature of the function affect the differentiability? 4. Could we say anything about the continuity of $$f(x)$$ at $$x = 0$$? 5. What are the geometric implications of the differing derivatives from the left and right? **Tip:** Always check continuity first before checking differentiability at a point in piecewise functions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differentiability
Piecewise Functions
Limits
Formulas
f'(x) = lim(h → 0) [f(x+h) - f(x)] / h
Right-hand limit: lim(h → 0^+) [f(h) - f(0)] / h
Left-hand limit: lim(h → 0^-) [f(h) - f(0)] / h
Theorems
Limit Definition of Derivative
Suitable Grade Level
Undergraduate Calculus (Math 160 level)
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