To identify all points at which $f(x)$ has an extremal value, we need to:

1. Compute the derivative $f'(x)$:

   = 3x^2 + 51x + 210.$$

2. Set $f'(x) = 0$ to find critical points: $3x^2 + 51x + 210 = 0.$

3. Solve the quadratic equation: Divide through by 3: $x^2 + 17x + 70 = 0.$ Factorize: $(x + 7)(x + 10) = 0.$ Thus, $x = -7$ and $x = -10$.

4. Determine the nature of these critical points (minimum, maximum, or saddle point): Compute the second derivative: $f''(x) = \frac{d}{dx}(3x^2 + 51x + 210) = 6x + 51.$ Evaluate $f''(x)$ at each critical point:

   • At $x = -7$: $f''(-7) = 6(-7) + 51 = -42 + 51 = 9 > 0 \quad (\text{local minimum}).$
   • At $x = -10$: $f''(-10) = 6(-10) + 51 = -60 + 51 = -9 < 0 \quad (\text{local maximum}).$

Conclusion:

• $x = -7$: Local minimum.
• $x = -10$: Local maximum.

Would you like me to compute the corresponding function values or delve deeper into graph analysis?

Related Questions:

1. How do second derivatives help classify extrema?
2. What is the significance of critical points in real-world applications?
3. Can this function have an absolute extremum? Why or why not?
4. How would the graph look around $x = -7$ and $x = -10$?
5. What happens if we remove the quadratic term from $f(x)$?

Tip: Always verify extrema by checking both the first and second derivatives.