https://file.aiquickdraw.com/mathbot/file/1728847669567o1bts9gv.jpg

The problem in the image asks to show that the piecewise function:

\[
f(x) = 
\begin{cases} 
x^2 + 2, & \text{if } x \leq 1 \\
x + 2, & \text{if } x > 1
\end{cases}
\]
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?

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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.

Show that f(x) = {x^2 + 2, x ≤ 1; x + 2, x > 1} is continuous but not differentiable at x = 1. Sketch the graph of f.

This problem involves proving that the piecewise function f(x) is continuous but not differentiable at x = 1. The solution checks the left-hand and right-hand limits for continuity and the left-hand and right-hand derivatives for differentiability. It includes a sketch of the graph of f.

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