Math Problem Statement
A function f ( x ) is given below. Which of the following is true for f ( x ) ? f ( x )
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ x x < 0 0 x
0 sin x 0 < x < π 2 1 x ≥ π 2
I. f ( x ) is continuous and differentiable at x
0 . II. f ( x ) is differentiable at x
π 2 . III. f ( x ) is continuous at x
0 and at x
π 2 .
Solution
Let's analyze the given function and the statements about its continuity and differentiability:
The function is defined piecewise as:
\begin{cases} x & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ \sin(x) & \text{if } 0 < x < \frac{\pi}{2} \\ 1 & \text{if } x \geq \frac{\pi}{2} \end{cases}$$ Now, let's assess each statement one by one: ### I. $$f(x)$$ is continuous and differentiable at $$x = 0$$. #### Continuity at $$x = 0$$: - To check continuity at $$x = 0$$, we need to see if the left-hand limit and right-hand limit exist and match the value of the function at $$x = 0$$. - **Left-hand limit** as $$x \to 0^-$$: When $$x < 0$$, $$f(x) = x$$. So, $$\lim_{x \to 0^-} f(x) = 0$$. - **Right-hand limit** as $$x \to 0^+$$: When $$0 < x < \frac{\pi}{2}$$, $$f(x) = \sin(x)$$. So, $$\lim_{x \to 0^+} f(x) = \sin(0) = 0$$. - Since $$f(0) = 0$$, and both the left-hand and right-hand limits are equal to $$0$$, the function is **continuous at $$x = 0$$**. #### Differentiability at $$x = 0$$: - To check differentiability at $$x = 0$$, we compute the derivative from both sides. - **Left-hand derivative**: $$\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h - 0}{h} = 1$$. - **Right-hand derivative**: $$\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\sin(h) - 0}{h} = \lim_{h \to 0^+} \frac{\sin(h)}{h} = 1$$. - Since both derivatives are equal to $$1$$, $$f(x)$$ is **differentiable at $$x = 0$$**. Thus, **Statement I is true**. --- ### II. $$f(x)$$ is differentiable at $$x = \frac{\pi}{2}$$. #### Continuity at $$x = \frac{\pi}{2}$$: - **Left-hand limit**: As $$x \to \frac{\pi}{2}^-$$, $$f(x) = \sin(x)$$. So, $$\lim_{x \to \frac{\pi}{2}^-} f(x) = \sin\left(\frac{\pi}{2}\right) = 1$$. - **Right-hand limit**: As $$x \to \frac{\pi}{2}^+$$, $$f(x) = 1$$. So, $$\lim_{x \to \frac{\pi}{2}^+} f(x) = 1$$. - Since both limits are equal to $$1$$, the function is **continuous at $$x = \frac{\pi}{2}$$**. #### Differentiability at $$x = \frac{\pi}{2}$$: - **Left-hand derivative**: $$\lim_{h \to 0^-} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} = \lim_{h \to 0^-} \frac{\sin\left(\frac{\pi}{2} + h\right) - 1}{h}$$. This gives $$\cos\left(\frac{\pi}{2}\right) = 0$$, so the derivative is $$0$$. - **Right-hand derivative**: $$\lim_{h \to 0^+} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} = \lim_{h \to 0^+} \frac{1 - 1}{h} = 0$$. - Since both derivatives are equal to $$0$$, $$f(x)$$ is **differentiable at $$x = \frac{\pi}{2}$$**. Thus, **Statement II is true**. --- ### III. $$f(x)$$ is continuous at $$x = 0$$ and at $$x = \frac{\pi}{2}$$. We have already checked that $$f(x)$$ is continuous at both $$x = 0$$ and $$x = \frac{\pi}{2}$$, so **Statement III is true**. --- ### Conclusion: All three statements (I, II, and III) are true. Would you like further clarification or more detailed explanations? Here are five related questions to explore: 1. What are the conditions for a function to be continuous at a point? 2. How do we determine if a function is differentiable at a point? 3. How does the derivative of piecewise functions behave at boundary points? 4. Can a function be continuous but not differentiable? Provide examples. 5. What are the key differences between differentiability and continuity? **Tip:** For piecewise functions, always check both continuity and differentiability separately at the transition points where the function definition changes.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Differentiability
Trigonometric Functions
Formulas
Limit Definition of Derivative
Sine Function
Continuity Condition (left-hand and right-hand limits)
Theorems
Continuity Theorem
Differentiability Theorem
Suitable Grade Level
Grades 11-12
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