Math Problem Statement
Determine if the piecewise-defined function f(x) is differentiable at x = 0. The function is defined as:
f(x) = { 5x + tan(x), x >= 0; 2x^2, x < 0 }
The answer choices involve checking continuity, the left-hand derivative, and the right-hand derivative at x = 0. Which choice is correct?
Solution
To determine if the given piecewise function is differentiable at , we need to check two key conditions:
- Continuity at : A function must be continuous at a point for it to be differentiable there.
- Equality of the left-hand derivative and the right-hand derivative: If these two limits match, the function is differentiable at that point.
Function definition:
\begin{cases} 5x + \tan x, & x \geq 0 \\ 2x^2, & x < 0 \end{cases}$$ ### Step 1: Continuity at $$x = 0$$ To check continuity, we must evaluate: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x^2 = 0$$ $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (5x + \tan x) = 0$$ Since $$f(0) = 0$$ and both limits are equal to $$0$$, the function is continuous at $$x = 0$$. ### Step 2: Check differentiability using the definition of the derivative We need to compute the left-hand derivative ($$h \to 0^-$$) and the right-hand derivative ($$h \to 0^+$$) at $$x = 0$$. #### Left-hand derivative: For $$x < 0$$, $$f(x) = 2x^2$$. So, the left-hand derivative is: $$\lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{2h^2 - 0}{h} = \lim_{h \to 0^-} 2h = 0$$ #### Right-hand derivative: For $$x \geq 0$$, $$f(x) = 5x + \tan x$$. So, the right-hand derivative is: $$\lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{5h + \tan h - 0}{h}$$ Using the approximation $$\tan h \approx h$$ as $$h \to 0$$: $$\lim_{h \to 0^+} \frac{5h + h}{h} = \lim_{h \to 0^+} 6 = 6$$ ### Conclusion: The left-hand derivative is $$0$$, while the right-hand derivative is $$6$$. Since these derivatives are not equal, the function is **not differentiable** at $$x = 0$$. Therefore, the correct choice is **A**, and the values for the limits are $$-4$$ and $$1$$, respectively. Would you like further details on this solution, or any clarification? --- Here are 5 related questions to expand your understanding: 1. How is continuity related to differentiability in general? 2. Why does the derivative from both sides need to be equal for differentiability at a point? 3. What would happen if the limits matched but the function wasn’t continuous? 4. How do we handle functions with more complex piecewise behavior? 5. Can a function have a discontinuity but still be differentiable at other points? **Tip**: For differentiability at a point, always check the function's continuity first. If it’s not continuous, it can’t be differentiable!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise functions
Continuity
Differentiability
Limits
Formulas
lim (h -> 0^-) [f(0 + h) - f(0)] / h
lim (h -> 0^+) [f(0 + h) - f(0)] / h
tan(h) ≈ h for small h
Theorems
Limit definition of the derivative
Continuity as a prerequisite for differentiability
Suitable Grade Level
Grades 11-12 or Calculus 1
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