Math Problem Statement
Solution
The problem in the image asks to show that the piecewise function:
\begin{cases} x^2 + 2, & \text{if } x \leq 1 \\ x + 2, & \text{if } x > 1 \end{cases}$$ is continuous but not differentiable at $$x = 1$$, and to sketch the graph of $$f$$. ### Step 1: Check Continuity at $$x = 1$$ For a function to be continuous at $$x = 1$$, the left-hand limit, right-hand limit, and the function value at $$x = 1$$ must all be equal. - Left-hand limit ($$\lim_{x \to 1^-} f(x)$$): When $$x \leq 1$$, $$f(x) = x^2 + 2$$. So, $$\lim_{x \to 1^-} f(x) = 1^2 + 2 = 3.$$ - Right-hand limit ($$\lim_{x \to 1^+} f(x)$$): When $$x > 1$$, $$f(x) = x + 2$$. So, $$\lim_{x \to 1^+} f(x) = 1 + 2 = 3.$$ - Function value at $$x = 1$$: $$f(1) = 1^2 + 2 = 3.$$ Since $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 3$$, the function is continuous at $$x = 1$$. ### Step 2: Check Differentiability at $$x = 1$$ For a function to be differentiable at $$x = 1$$, the left-hand derivative and right-hand derivative must be equal at $$x = 1$$. - Left-hand derivative ($$\lim_{x \to 1^-} f'(x)$$): The derivative of $$f(x) = x^2 + 2$$ is $$f'(x) = 2x$$. So, $$\lim_{x \to 1^-} f'(x) = 2 \cdot 1 = 2.$$ - Right-hand derivative ($$\lim_{x \to 1^+} f'(x)$$): The derivative of $$f(x) = x + 2$$ is $$f'(x) = 1$$. So, $$\lim_{x \to 1^+} f'(x) = 1.$$ Since $$\lim_{x \to 1^-} f'(x) = 2$$ and $$\lim_{x \to 1^+} f'(x) = 1$$, the left-hand and right-hand derivatives are not equal. Therefore, the function is **not differentiable** at $$x = 1$$. ### Step 3: Sketch the Graph of $$f(x)$$ - For $$x \leq 1$$, the function follows $$f(x) = x^2 + 2$$, which is a parabola shifted upward by 2. - For $$x > 1$$, the function follows $$f(x) = x + 2$$, which is a straight line with slope 1 and intercept 2. - The function is continuous at $$x = 1$$, meaning there is no break in the graph, but there is a corner or sharp point at $$x = 1$$ due to the different derivatives on each side. Would you like more details on any part, or have any questions? --- Here are 5 related questions you might explore: 1. What is the formal definition of continuity and differentiability for a function? 2. How can one use the definition of a derivative to check differentiability at a point? 3. Can you provide other examples of functions that are continuous but not differentiable at certain points? 4. How does the geometric shape of a function's graph help in identifying continuity and differentiability? 5. What role does piecewise definition play in determining the differentiability of functions? **Tip**: When checking for differentiability, always ensure to verify both the left-hand and right-hand derivatives.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Differentiability
Piecewise Functions
Formulas
Left-hand limit: lim(x -> 1^-) f(x)
Right-hand limit: lim(x -> 1^+) f(x)
Derivative: f'(x)
Theorems
A function is continuous at x = a if lim(x -> a^-) f(x) = lim(x -> a^+) f(x) = f(a).
A function is differentiable at x = a if the left-hand and right-hand derivatives are equal at a.
Suitable Grade Level
Grades 11-12, Early College
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