Differentiability and Continuity of Piecewise Functions at x = 1 and x = -1

Math Problem Statement

Problem 1: Show that f(x) is differentiable at x = 1 and compute f'(1). Problem 2: Show that f(x) is continuous at x = -1 and that f(x) is not differentiable at x = -1.

Solution

Let's break down and solve the problems.

Problem 1: Differentiability at $x = 1$

The function is defined as:

\begin{cases} x^3 & \text{if } x \leq 1 \\ 3x - 2 & \text{if } x > 1 \end{cases}$$ **Step 1: Check if $$f(x)$$ is continuous at $$x = 1$$** For $$f(x)$$ to be differentiable at $$x = 1$$, it must first be continuous at $$x = 1$$. We need to check if the left-hand limit and right-hand limit at $$x = 1$$ are equal, and if they match the function value at $$x = 1$$. - Left-hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^3 = 1^3 = 1$$ - Right-hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3x - 2) = 3(1) - 2 = 1$$ - Function value: $$f(1) = 1^3 = 1$$ Since the left-hand limit, right-hand limit, and the function value are all equal to 1, $$f(x)$$ is continuous at $$x = 1$$. **Step 2: Check if $$f(x)$$ is differentiable at $$x = 1$$** Next, we check the derivative from both sides at $$x = 1$$. - Left-hand derivative: $$f'(x) = \frac{d}{dx} (x^3) = 3x^2$$ $$f'(1^-) = 3(1)^2 = 3$$ - Right-hand derivative: $$f'(x) = \frac{d}{dx} (3x - 2) = 3$$ $$f'(1^+) = 3$$ Since the left-hand derivative and the right-hand derivative are both equal to 3, $$f(x)$$ is differentiable at $$x = 1$$. **Conclusion for Problem 1:** - $$f(x)$$ is differentiable at $$x = 1$$. - The derivative at $$x = 1$$ is $$f'(1) = 3$$. --- ### Problem 2: Continuity and Non-differentiability at $$x = -1$$ The function is defined as: $$f(x) = \begin{cases} x^2 & \text{if } x \leq -1 \\ 1 & \text{if } x > -1 \end{cases}$$ **Step 1: Check if $$f(x)$$ is continuous at $$x = -1$$** For continuity at $$x = -1$$, we need the left-hand limit and right-hand limit to be equal, and to match the function value. - Left-hand limit: $$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} x^2 = (-1)^2 = 1$$ - Right-hand limit: $$\lim_{x \to -1^+} f(x) = 1$$ - Function value: $$f(-1) = (-1)^2 = 1$$ Since the left-hand limit, right-hand limit, and function value are all equal to 1, $$f(x)$$ is continuous at $$x = -1$$. **Step 2: Check if $$f(x)$$ is differentiable at $$x = -1$$** Now, let's check the derivative from both sides. - Left-hand derivative: $$f'(x) = \frac{d}{dx} (x^2) = 2x$$ $$f'(-1^-) = 2(-1) = -2$$ - Right-hand derivative: For $$x > -1$$, $$f(x) = 1$$, which is constant, so the derivative is 0. $$f'(-1^+) = 0$$ Since the left-hand derivative and right-hand derivative are not equal (one is -2, the other is 0), $$f(x)$$ is **not differentiable** at $$x = -1$$. **Conclusion for Problem 2:** - $$f(x)$$ is continuous at $$x = -1$$. - $$f(x)$$ is not differentiable at $$x = -1$$. --- Do you have any questions or would you like more details on any part? Here are 5 related questions for further exploration: 1. How do the conditions for continuity and differentiability differ? 2. Why does a function being continuous not necessarily imply it is differentiable? 3. Can a function be differentiable at a point but not continuous? 4. How can we use limits to check for differentiability? 5. What are some typical cases where a function is continuous but not differentiable? **Tip**: Remember, differentiability implies continuity, but continuity alone does not imply differentiability!

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Math Problem Analysis

Mathematical Concepts

Differentiability
Continuity
Piecewise Functions
Limits

Formulas

Limit definition of derivative: f'(x) = lim(h -> 0) [(f(x + h) - f(x)) / h]
Left-hand and right-hand limits for continuity
Piecewise function evaluation

Theorems

Differentiability implies continuity
Conditions for continuity at a point
Conditions for differentiability at a point

Suitable Grade Level

College/University Calculus

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