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.