Math Problem Statement
Let \( f \) be a function defined by \( f(x) = \begin{cases} 1 - 2 \sin x & \text{for } x \leq 0 \\ e^{-4x} & \text{for } x > 0 \end{cases} \).
(a) Show that \( f \) is continuous at \( x = 0 \).
(b) For \( x \neq 0 \), express \( f'(x) \) as a piecewise-defined function. Find the value of \( x \) for which \( f'(x) = -3 \).
Solution
Let's solve the problem step-by-step.
Given:
A function defined piecewise by:
1 - 2 \sin x & \text{for } x \leq 0 \\ e^{-4x} & \text{for } x > 0 \end{cases}$$ We need to address parts (a) and (b). --- ### Part (a): Show that $$f$$ is continuous at $$x = 0$$. To prove continuity at $$x = 0$$, we need to show that: 1. $$f(0)$$ exists. 2. $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)$$. 3. $$\lim_{x \to 0} f(x) = f(0)$$. **Step 1: Evaluate $$f(0)$$:** Since $$f(x) = 1 - 2 \sin x$$ for $$x \leq 0$$, $$f(0) = 1 - 2 \sin 0 = 1 - 2 \cdot 0 = 1.$$ **Step 2: Find $$\lim_{x \to 0^-} f(x)$$:** For $$x \leq 0$$, $$f(x) = 1 - 2 \sin x$$. $$\lim_{x \to 0^-} f(x) = 1 - 2 \sin 0 = 1.$$ **Step 3: Find $$\lim_{x \to 0^+} f(x)$$:** For $$x > 0$$, $$f(x) = e^{-4x}$$. $$\lim_{x \to 0^+} f(x) = e^{-4 \cdot 0} = e^0 = 1.$$ Since $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1$$, the function $$f(x)$$ is continuous at $$x = 0$$. --- ### Part (b): For $$x \neq 0$$, express $$f'(x)$$ as a piecewise-defined function. Find the value of $$x$$ for which $$f'(x) = -3$$. To find $$f'(x)$$, we differentiate each piece separately. 1. **For $$x \leq 0$$:** $$f(x) = 1 - 2 \sin x$$ $$f'(x) = -2 \cos x.$$ 2. **For $$x > 0$$:** $$f(x) = e^{-4x}$$ $$f'(x) = -4 e^{-4x}.$$ Thus, we can write $$f'(x)$$ as: $$f'(x) = \begin{cases} -2 \cos x & \text{for } x \leq 0 \\ -4 e^{-4x} & \text{for } x > 0 \end{cases}$$ **Finding $$x$$ for which $$f'(x) = -3$$:** We need to solve $$f'(x) = -3$$ for both cases. 1. **For $$x \leq 0$$:** $$-2 \cos x = -3 \Rightarrow \cos x = \frac{3}{2}.$$ This equation has no solutions because $$\cos x$$ ranges only from -1 to 1. 2. **For $$x > 0$$:** $$-4 e^{-4x} = -3 \Rightarrow e^{-4x} = \frac{3}{4}.$$ Taking the natural logarithm on both sides: $$-4x = \ln \frac{3}{4} \Rightarrow x = -\frac{1}{4} \ln \frac{3}{4}.$$ So, the solution for $$x$$ where $$f'(x) = -3$$ is: $$x = -\frac{1}{4} \ln \frac{3}{4}.$$ --- ### Summary of Solutions: - **(a)** $$f(x)$$ is continuous at $$x = 0$$. - **(b)** $$f'(x)$$ as a piecewise-defined function: $$f'(x) = \begin{cases} -2 \cos x & \text{for } x \leq 0 \\ -4 e^{-4x} & \text{for } x > 0 \end{cases}$$ The value of $$x$$ for which $$f'(x) = -3$$ is: $$x = -\frac{1}{4} \ln \frac{3}{4}.$$ Would you like further details or have any questions? --- ### Related Questions 1. How would the answer change if the function were defined differently for $$x > 0$$? 2. Can a discontinuous function ever be differentiable? Why or why not? 3. How does the concept of limits play a role in determining continuity? 4. What are the general steps to determine the continuity of a piecewise function? 5. How can you find higher-order derivatives for piecewise functions like $$f(x)$$? #### Tip When working with piecewise functions, always check continuity at the points where the pieces meet. This often involves calculating left-hand and right-hand limits.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Continuity
Differentiation
Piecewise Functions
Formulas
f(x) = \begin{cases} 1 - 2 \sin x & \text{for } x \leq 0 \\ e^{-4x} & \text{for } x > 0 \end{cases}
Continuity: \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)
Differentiation of piecewise functions
Theorems
Continuity at a point
Derivative of trigonometric and exponential functions
Suitable Grade Level
Grades 11-12, College Calculus
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