Math Problem Statement
Solution
The question involves discussing the continuity and differentiability of the given piecewise function:
\begin{cases} 1 - x, & x < 1 \\ (1 - x)(2 - x), & 1 \leq x \leq 2 \\ 3 - x, & x > 2 \end{cases}$$ ### Step 1: **Check continuity at $$x = 1$$ and $$x = 2$$** Continuity requires that: $$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a),$$ where $$a$$ is the point of interest. #### At $$x = 1$$: - Left-hand limit ($$x \to 1^-$$): Using $$f(x) = 1 - x$$, $$\lim_{x \to 1^-} f(x) = 1 - 1 = 0.$$ - Right-hand limit ($$x \to 1^+$$): Using $$f(x) = (1 - x)(2 - x)$$, $$\lim_{x \to 1^+} f(x) = (1 - 1)(2 - 1) = 0.$$ - Value of $$f(1)$$: Using $$f(x) = (1 - x)(2 - x)$$, $$f(1) = (1 - 1)(2 - 1) = 0.$$ Thus, $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 0$$, so $$f(x)$$ is continuous at $$x = 1$$. #### At $$x = 2$$: - Left-hand limit ($$x \to 2^-$$): Using $$f(x) = (1 - x)(2 - x)$$, $$\lim_{x \to 2^-} f(x) = (1 - 2)(2 - 2) = 0.$$ - Right-hand limit ($$x \to 2^+$$): Using $$f(x) = 3 - x$$, $$\lim_{x \to 2^+} f(x) = 3 - 2 = 1.$$ - Value of $$f(2)$$: Using $$f(x) = (1 - x)(2 - x)$$, $$f(2) = (1 - 2)(2 - 2) = 0.$$ Thus, $$\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$$, so $$f(x)$$ is **not continuous** at $$x = 2$$. ### Step 2: **Check differentiability** Differentiability implies continuity, so $$f(x)$$ is **not differentiable at $$x = 2$$** because it is not continuous there. #### At $$x = 1$$: Compute the derivatives for $$x < 1$$ and $$1 \leq x \leq 2$$, and check if they are equal at $$x = 1$$. - For $$x < 1$$: $$f(x) = 1 - x$$, so $$f'(x) = -1$$. - For $$1 \leq x \leq 2$$: $$f(x) = (1 - x)(2 - x) = 2 - 3x + x^2$$, so $$f'(x) = -3 + 2x.$$ At $$x = 1$$, $$f'(1) = -3 + 2(1) = -1.$$ The derivatives match ($$f'(x) = -1$$) at $$x = 1$$. Thus, $$f(x)$$ is differentiable at $$x = 1$$. ### Final Answer: - $$f(x)$$ is **continuous** and **differentiable** at $$x = 1$$. - $$f(x)$$ is **not continuous** and hence **not differentiable** at $$x = 2$$. --- Would you like detailed steps for derivative calculations or further exploration of this function? ### 5 Related Questions: 1. How do you formally prove differentiability for a piecewise function? 2. What is the geometric interpretation of continuity and differentiability? 3. How do limits relate to continuity in functions? 4. Can you calculate higher-order derivatives for $$f(x)$$ in each interval? 5. What happens if we redefine $$f(x)$$ at $$x = 2$$ to make it continuous? ### Tip: When working with piecewise functions, always check both continuity and differentiability at the boundaries by carefully evaluating the limits and derivatives from both sides.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Differentiability
Formulas
f(x) = 1 - x for x < 1
f(x) = (1 - x)(2 - x) for 1 ≤ x ≤ 2
f(x) = 3 - x for x > 2
Definition of continuity: lim(x → a-) f(x) = lim(x → a+) f(x) = f(a)
Definition of differentiability: matching derivatives from left and right at a point
Theorems
Continuity at a Point
Differentiability Implies Continuity
Suitable Grade Level
Grades 11-12
Related Recommendation
Differentiability and Continuity of Piecewise Functions at x = 1 and x = -1
Continuity and Differentiability of a Piecewise Function
Continuity and Differentiability of a Piecewise Function
Testing Continuity and Differentiability of a Piecewise Function at x = 1/2
Differentiability of a Piecewise Function at x = 3