Math Problem Statement
A function f x is defined as follows:
2
1 0 1 sin 0 2
2 2 2 when x f x x when x x when x
,
Discuss the differentiability at x 0 and 2 x .
Solution
To analyze the differentiability of the given piecewise function
\begin{cases} 1 & \text{if } x < 0, \\ 1 + \sin x & \text{if } 0 \leq x < \frac{\pi}{2}, \\ x^2 + 2x - 2 & \text{if } x \geq \frac{\pi}{2}, \end{cases}$$ at $$x = 0$$ and $$x = \frac{\pi}{2}$$, we will follow these steps: --- ### 1. Differentiability at $$x = 0$$ To determine differentiability at $$x = 0$$, we check: 1. **Continuity at $$x = 0$$:** - From the left ($$x < 0$$): $$f(x) = 1$$. $$\lim_{x \to 0^-} f(x) = 1.$$ - From the right ($$0 \leq x < \frac{\pi}{2}$$): $$f(x) = 1 + \sin x$$. $$\lim_{x \to 0^+} f(x) = 1 + \sin(0) = 1.$$ - The function value is $$f(0) = 1$$. Since $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$$, the function is continuous at $$x = 0$$. 2. **Differentiability at $$x = 0$$:** The derivative from the left is: $$f'(x) = 0 \quad \text{for } x < 0.$$ So, $$f'(0^-) = 0.$$ The derivative from the right is: $$f'(x) = \cos x \quad \text{for } 0 \leq x < \frac{\pi}{2}.$$ At $$x = 0$$, $$f'(0^+) = \cos(0) = 1.$$ Since $$f'(0^-) \neq f'(0^+)$$, the function is **not differentiable at $$x = 0$$**. --- ### 2. Differentiability at $$x = \frac{\pi}{2}$$ To determine differentiability at $$x = \frac{\pi}{2}$$, we check: 1. **Continuity at $$x = \frac{\pi}{2}$$:** - From the left ($$0 \leq x < \frac{\pi}{2}$$): $$f(x) = 1 + \sin x$$. $$\lim_{x \to \frac{\pi}{2}^-} f(x) = 1 + \sin\left(\frac{\pi}{2}\right) = 1 + 1 = 2.$$ - From the right ($$x \geq \frac{\pi}{2}$$): $$f(x) = x^2 + 2x - 2$$. $$\lim_{x \to \frac{\pi}{2}^+} f(x) = \left(\frac{\pi}{2}\right)^2 + 2\left(\frac{\pi}{2}\right) - 2.$$ Substituting: $$\lim_{x \to \frac{\pi}{2}^+} f(x) = \frac{\pi^2}{4} + \pi - 2.$$ Since $$\lim_{x \to \frac{\pi}{2}^-} f(x) \neq \lim_{x \to \frac{\pi}{2}^+} f(x)$$, the function is **not continuous at $$x = \frac{\pi}{2}$$**. 2. Since differentiability requires continuity, $$f(x)$$ is **not differentiable at $$x = \frac{\pi}{2}$$**. --- ### Conclusion: - $$f(x)$$ is **not differentiable at $$x = 0$$** because the derivative is not continuous. - $$f(x)$$ is **not differentiable at $$x = \frac{\pi}{2}$$** because the function is not even continuous there. --- Do you want detailed derivations or graphs for further understanding? --- ### Related Questions: 1. What are the conditions for differentiability of a piecewise function? 2. How does continuity relate to differentiability? 3. Can a function be differentiable at a point where its derivative is undefined? 4. What is the geometric interpretation of non-differentiability at a point? 5. How does the concept of left-hand and right-hand derivatives help analyze differentiability? --- ### Tip: Always verify continuity before testing differentiability. Discontinuity guarantees non-differentiability.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Differentiability
Continuity
Limits
Formulas
f'(x) = lim (h -> 0) [f(x+h) - f(x)] / h
Derivative of sin(x): d/dx[sin(x)] = cos(x)
Theorems
A function is differentiable at a point if it is continuous at that point and the left-hand derivative equals the right-hand derivative.
Continuity requires that lim(x -> c-) f(x) = lim(x -> c+) f(x) = f(c).
Suitable Grade Level
Undergraduate Calculus or Advanced High School
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