Math Problem Statement
If \( f: \mathbb{R} \to \mathbb{R} \) is defined by \( f(x) = \begin{cases} e^{x/2} - 1, & x \leq 0 \\ \frac{1 - \cos x}{x}, & x > 0 \end{cases} \), what is the value of \( f'(0) \)?
Solution
To solve this problem, we need to find the derivative of at , which is , given that:
\begin{cases} e^{x/2} - 1, & x \leq 0 \\ \frac{1 - \cos x}{x}, & x > 0 \end{cases}$$ ### Step-by-Step Solution 1. **Left-Hand Limit** (when $$x \to 0^-$$): Since $$x \leq 0$$ part of the function is $$f(x) = e^{x/2} - 1$$, let's calculate $$f'(0^-)$$: $$f'(x) = \frac{d}{dx}\left( e^{x/2} - 1 \right) = \frac{1}{2} e^{x/2}$$ Now, we calculate the limit as $$x \to 0^-$$: $$f'(0^-) = \frac{1}{2} e^{0/2} = \frac{1}{2}$$ 2. **Right-Hand Limit** (when $$x \to 0^+$$): For $$x > 0$$, $$f(x) = \frac{1 - \cos x}{x}$$. To find $$f'(0^+)$$, observe that as $$x \to 0$$, the function $$\frac{1 - \cos x}{x}$$ resembles the derivative of $$\sin^2(x/2) / x$$, which approaches zero. Thus, it is also $$\frac{1}{2}$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Differentiation
Piecewise Functions
Formulas
Derivative of \( e^{x/2} \)
Derivative of \( \frac{1 - \cos x}{x} \)
Theorems
Limit definition of derivative
Squeeze theorem
Suitable Grade Level
Undergraduate
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