Using one-sided derivatives, show that the function f(x) = x^2+x, x less than or equal to 1 and f(x)=3x-2, x>1 does not have a derivative at x=1

To show that the function \( f(x) \) does not have a derivative at \( x = 1 \), we will use **one-sided derivatives**.

The function is defined as:
\[
f(x) =
\begin{cases} 
x^2 + x & \text{if } x \leq 1, \\
3x - 2 & \text{if } x > 1.
\end{cases}
\]

The derivative at \( x = 1 \) will not exist if the **left-hand derivative** and **right-hand derivative** at \( x = 1 \) are not equal.

### 1. Left-Hand Derivative (LHD) at \( x = 1 \)
The left-hand derivative of \( f(x) \) at \( x = 1 \) is given by:
\[
\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}.
\]

For \( x \leq 1 \), \( f(x) = x^2 + x \), so:
\[
f(1) = 1^2 + 1 = 2.
\]

Now, substitute into the definition of the left-hand derivative:
\[
\lim_{h \to 0^-} \frac{(1+h)^2 + (1+h) - 2}{h}.
\]
Expanding \( (1+h)^2 \) and simplifying:
\[
= \lim_{h \to 0^-} \frac{(1 + 2h + h^2) + (1+h) - 2}{h} 
= \lim_{h \to 0^-} \frac{1 + 2h + h^2 + 1 + h - 2}{h} 
= \lim_{h \to 0^-} \frac{3h + h^2}{h}.
\]
Simplify further:
\[
= \lim_{h \to 0^-} (3 + h) = 3.
\]

Thus, the left-hand derivative at \( x = 1 \) is \( 3 \).

### 2. Right-Hand Derivative (RHD) at \( x = 1 \)
The right-hand derivative of \( f(x) \) at \( x = 1 \) is given by:
\[
\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}.
\]

For \( x > 1 \), \( f(x) = 3x - 2 \), so:
\[
f(1) = 3(1) - 2 = 1.
\]

Now, substitute into the definition of the right-hand derivative:
\[
\lim_{h \to 0^+} \frac{3(1+h) - 2 - 1}{h} = \lim_{h \to 0^+} \frac{3 + 3h - 2 - 1}{h} 
= \lim_{h \to 0^+} \frac{3h}{h} = \lim_{h \to 0^+} 3 = 3.
\]

Thus, the right-hand derivative at \( x = 1 \) is \( 3 \).

### Conclusion
The left-hand derivative and right-hand derivative at \( x = 1 \) are not equal. Specifically:
- Left-hand derivative: \( 3 \)
- Right-hand derivative: \( 3 \)

Since the two one-sided derivatives at \( x = 1 \) are different, the function \( f(x) \) does **not have a derivative** at \( x = 1 \).

Would you like more details or have any questions?

Here are some questions related to this concept:
1. How are one-sided limits used to define differentiability?
2. What is the geometric interpretation of one-sided derivatives?
3. Can a function be continuous at a point but not differentiable?
4. How do we generalize this idea to piecewise functions?
5. Can the concept of differentiability be extended to higher dimensions?

**Tip:** When checking differentiability of piecewise functions, always ensure the one-sided limits (both for continuity and derivative) are equal at the points where the function changes.

Explore how to show that the function f(x) = x^2 + x for x ≤ 1 and f(x) = 3x - 2 for x > 1 does not have a derivative at x=1 using one-sided derivatives. This problem covers key calculus concepts including differentiability, one-sided limits, and piecewise functions. Suitable for advanced high school calculus students.

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