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To compute \( f'(-1) \) using the definition of the derivative, we start with the function:
\[
f(x) = 2x^2 + 2x + 3
\]

### Step 1: Definition of the Derivative
The derivative of \( f(x) \) at \( x = -1 \) is given by:
\[
f'(-1) = \lim_{h \to 0} \frac{f(-1 + h) - f(-1)}{h}
\]

### Step 2: Compute \( f(-1) \)
Substitute \( x = -1 \) into \( f(x) \):
\[
f(-1) = 2(-1)^2 + 2(-1) + 3 = 2 \cdot 1 - 2 + 3 = 3
\]

### Step 3: Compute \( f(-1 + h) \)
Substitute \( x = -1 + h \) into \( f(x) \):
\[
f(-1 + h) = 2(-1 + h)^2 + 2(-1 + h) + 3
\]

Expanding \( (-1 + h)^2 \):
\[
(-1 + h)^2 = 1 - 2h + h^2
\]
Then,
\[
f(-1 + h) = 2(1 - 2h + h^2) + 2(-1 + h) + 3
\]
\[
= 2 - 4h + 2h^2 - 2 + 2h + 3
\]
\[
= 2h^2 - 2h + 3
\]

### Step 4: Substitute into the Difference Quotient
Now, substitute \( f(-1 + h) \) and \( f(-1) \) into the difference quotient:
\[
f'(-1) = \lim_{h \to 0} \frac{(2h^2 - 2h + 3) - 3}{h}
\]
\[
= \lim_{h \to 0} \frac{2h^2 - 2h}{h}
\]
\[
= \lim_{h \to 0} \frac{h(2h - 2)}{h}
\]
\[
= \lim_{h \to 0} (2h - 2) = -2
\]

### Conclusion
Therefore, \( f'(-1) = -2 \).

Would you like more details or have questions on any step?

---

1. How is the derivative computed for different polynomial functions?
2. What are other methods to find the derivative without limits?
3. Why is the limit definition of the derivative important?
4. Can we generalize the derivative formula for quadratic functions?
5. How do we interpret \( f'(-1) = -2 \) in a real-world context?

**Tip:** Practicing the limit definition of the derivative helps solidify your understanding of the derivative's concept beyond just using rules.

Use the definition of the derivative to compute f'(-1) if f(x) = 2x^2 + 2x + 3.

Find the derivative f'(-1) for the function f(x) = 2x^2 + 2x + 3 using the limit definition of the derivative. This solution involves calculus concepts such as limits and derivatives, suitable for Grade 12 or first-year college students.

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