To find the local minimum and maximum of the function $f(x) = -2x^3 + 42x^2 - 270x + 1$, we need to determine the critical points where $f'(x) = 0$ and then analyze the second derivative to identify the nature of each critical point.

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

   $f(x) = -2x^3 + 42x^2 - 270x + 1$ Differentiating with respect to $x$: $f'(x) = -6x^2 + 84x - 270$

2. Set $f'(x) = 0$ to find critical points:

   $-6x^2 + 84x - 270 = 0$

   Simplifying this equation by dividing by -6: $x^2 - 14x + 45 = 0$

   Now we can solve this quadratic equation using the quadratic formula: $x = \frac{14 \pm \sqrt{14^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1}$ $x = \frac{14 \pm \sqrt{196 - 180}}{2}$ $x = \frac{14 \pm \sqrt{16}}{2}$ $x = \frac{14 \pm 4}{2}$

   This yields: $x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{10}{2} = 5$

   So, the critical points are $x = 5$ and $x = 9$.

3. Determine the nature of each critical point by evaluating $f''(x)$:

   Compute the second derivative $f''(x)$: $f'(x) = -6x^2 + 84x - 270$ Differentiating again: $f''(x) = -12x + 84$

   Now evaluate $f''(x)$ at each critical point.

   • For $x = 5$: $f''(5) = -12(5) + 84 = -60 + 84 = 24$ Since $f''(5) > 0$, this indicates a local minimum at $x = 5$.

   • For $x = 9$: $f''(9) = -12(9) + 84 = -108 + 84 = -24$ Since $f''(9) < 0$, this indicates a local maximum at $x = 9$.

4. Calculate the function values at these points:

   • Local minimum at $x = 5$: $f(5) = -2(5)^3 + 42(5)^2 - 270(5) + 1$ $= -2(125) + 42(25) - 1350 + 1$ $= -250 + 1050 - 1350 + 1 = -549$

   • Local maximum at $x = 9$: $f(9) = -2(9)^3 + 42(9)^2 - 270(9) + 1$ $= -2(729) + 42(81) - 2430 + 1$ $= -1458 + 3402 - 2430 + 1 = 515$

Final Answer:

• The function has a local minimum at $x = 5$ with value $-549$.
• The function has a local maximum at $x = 9$ with value $515$.

Would you like further details or have any questions?

Related Questions

1. How does the second derivative test help in identifying local extrema?
2. What would happen if $f(x)$ had a second derivative that was zero at a critical point?
3. Can you find the inflection points of this function?
4. How does the shape of $f(x) = -2x^3 + 42x^2 - 270x + 1$ look like?
5. How do critical points relate to the concavity of a function?

Tip

Remember that for polynomials, local extrema often correspond to the changes in the sign of the first derivative, making them easier to identify visually by looking at the graph.