https://file.aiquickdraw.com/mathbot/file/1730176683776xwf5pr23.jpg

Let's analyze the function \( f(x) \) given in the image:

\[
f(x) = \begin{cases} 
x^2 - 6, & x < 0 \\
-6, & x \geq 0 
\end{cases}
\]

### Part (A): Sketch the Graph
The function has two parts:
1. For \( x < 0 \), \( f(x) = x^2 - 6 \), which is a parabola shifted down by 6 units.
2. For \( x \geq 0 \), \( f(x) = -6 \), which is a constant function.

To sketch:
- For \( x < 0 \), the parabola \( f(x) = x^2 - 6 \) opens upwards, and its vertex is at \( (0, -6) \).
- For \( x \geq 0 \), the graph is a horizontal line at \( f(x) = -6 \).

### Part (B): Continuity Check
To check for continuity, observe what happens at \( x = 0 \):
1. The left-hand limit as \( x \to 0^- \) is \( \lim_{x \to 0^-} f(x) = 0^2 - 6 = -6 \).
2. The right-hand limit as \( x \to 0^+ \) is \( \lim_{x \to 0^+} f(x) = -6 \).

Since the left-hand limit, right-hand limit, and the function value at \( x = 0 \) all equal \( -6 \), the function is **continuous** at \( x = 0 \). Therefore, **no points of discontinuity exist** for \( f(x) \).

Answer for Part (B): **NONE**

### Part (C): Differentiability Check
To check differentiability at \( x = 0 \):
1. Compute the derivative for \( x < 0 \): \( f'(x) = 2x \).
   - As \( x \to 0^- \), \( f'(x) \to 0 \).
2. For \( x \geq 0 \), \( f(x) = -6 \), so \( f'(x) = 0 \) for \( x > 0 \).

However, at \( x = 0 \), there is a **cusp** due to the abrupt change from a quadratic function to a constant function. Hence, \( f \) is **not differentiable** at \( x = 0 \).

Answer for Part (C): \( x = 0 \)

Would you like further details on any part? 

---

### Follow-Up Questions:
1. How is differentiability related to continuity at a point?
2. What are common indicators of a function being non-differentiable at a point?
3. Why does a cusp cause non-differentiability?
4. How can we verify continuity for piecewise functions in general?
5. What other examples of functions have similar behavior to this one at a boundary?

### Tip:
For piecewise functions, checking continuity and differentiability at transition points is essential to understand their behavior fully.

Let \( f(x) = \begin{cases} x^2 - 6, & x < 0 \\ -6, & x \geq 0 \end{cases} \). (A) Sketch the graph of \( f \). (B) Find the value of \( x \) where \( f \) is discontinuous. (C) Find the value of \( x \) where \( f \) is not differentiable.

Explore a problem on continuity and differentiability of a piecewise function \( f(x) = \begin{cases} x^2 - 6, & x < 0 \\ -6, & x \geq 0 \end{cases} \). Learn how to determine continuity and differentiability at transition points for a piecewise-defined function, suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation