Consider the function $f(x)$ defined piecewise as follows:

$\[

f(x) = 

\begin{cases} 

x^2 & \text{if } x < 1 \\ 

2x + 1 & \text{if } 1 \leq x < 3 \\ 

4 & \text{if } x \geq 3 

\end{cases}

\]$

Evaluate the limit

$\[

\lim_{x \to 1^-} f'(x)

\]$

and 

$\[

\lim_{x \to 1^+} f'(x)

\]$

Based on your findings, is $f(x)$ differentiable at $x=1$? Explain why or why not.

We are tasked with evaluating the limits of the derivative of the piecewise function $ f(x) $ as $ x $ approaches 1 from the left ($ x \to 1^- $) and from the right ($ x \to 1^+ $). We will also determine whether $ f(x) $ is differentiable at $ x = 1 $.

### Step 1: Evaluate $ \lim_{x \to 1^-} f'(x) $

For $ x < 1 $, the function is given by $ f(x) = x^2 $. 

The derivative of $ f(x) = x^2 $ is:
\[
f'(x) = 2x
\]
Thus, as $ x \to 1^- $:
\[
\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} 2x = 2(1) = 2
\]

### Step 2: Evaluate $ \lim_{x \to 1^+} f'(x) $

For $ 1 \leq x < 3 $, the function is given by $ f(x) = 2x + 1 $.

The derivative of $ f(x) = 2x + 1 $ is:
\[
f'(x) = 2
\]
Thus, as $ x \to 1^+ $:
\[
\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} 2 = 2
\]

### Step 3: Determine whether $ f(x) $ is differentiable at $ x = 1 $

To determine if $ f(x) $ is differentiable at $ x = 1 $, the following conditions must hold:
1. $ f(x) $ must be continuous at $ x = 1 $.
2. The derivative from the left must equal the derivative from the right, i.e., $ \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) $.

#### Continuity at $ x = 1 $:

We check the value of $ f(x) $ at $ x = 1 $. From the piecewise definition:
- For $ x < 1 $, $ f(x) = x^2 $, so $ f(1^-) = 1^2 = 1 $.
- For $ x \geq 1 $, $ f(x) = 2x + 1 $, so $ f(1^+) = 2(1) + 1 = 3 $.

Since $ f(1^-) = 1 $ and $ f(1^+) = 3 $, the function $ f(x) $ is **not continuous** at $ x = 1 $.

#### Derivatives at $ x = 1 $:

We already found:
\[
\lim_{x \to 1^-} f'(x) = 2 \quad \text{and} \quad \lim_{x \to 1^+} f'(x) = 2
\]
Since these two limits are equal, the derivative from the left and right matches.

### Conclusion:

Even though the derivative from the left and right is the same ($ f'(x) = 2 $ as $ x \to 1 $), the function is **not continuous at $ x = 1 $**. A function must be continuous at a point to be differentiable there. Hence, $ f(x) $ is **not differentiable at $ x = 1 $** because it is not continuous at that point.

---

Would you like further details on any step or have any questions? Here are some related questions to explore:

1. What is the formal definition of differentiability at a point?
2. How does continuity relate to differentiability?
3. Can a function have a continuous derivative but still be non-differentiable at a point?
4. What are other conditions that can prevent differentiability at a point?
5. How does the piecewise nature of a function affect its differentiability?

**Tip:** Always check both continuity and the equality of derivatives from both sides when verifying differentiability at a point.

Consider the function $f(x)$ defined piecewise as follows:

$f(x) = 
\begin{cases} 
 x^2 & \text{if } x < 1 \
 2x + 1 & \text{if } 1 \leq x < 3 \
 4 & \text{if } x \geq 3 
\end{cases}$

Evaluate the limit

$\lim_{x \to 1^-} f'(x)$

and

$\lim_{x \to 1^+} f'(x)$

Based on your findings, is $f(x)$ differentiable at $x=1$? Explain why or why not.

Explore how to evaluate the left-hand and right-hand derivatives of a piecewise function at x=1 and determine its differentiability. This problem involves key concepts like limits, continuity, and the conditions for differentiability. Ideal for students in Grades 11-12.

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