The function given is $f(x) = 2x^3 - 39x^2 + 180x + 9$. To estimate the local minimum and maximum, we need to find where the derivative of the function equals zero and determine the nature of those points.

Step 1: Find the first derivative

Differentiate $f(x)$ with respect to $x$:

$f'(x) = 6x^2 - 78x + 180$

Step 2: Find the critical points

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

$6x^2 - 78x + 180 = 0$

Divide the entire equation by 6:

$x^2 - 13x + 30 = 0$

Solve this quadratic equation using the quadratic formula:

$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(1)(30)}}{2(1)}$ $x = \frac{13 \pm \sqrt{169 - 120}}{2}$ $x = \frac{13 \pm \sqrt{49}}{2}$ $x = \frac{13 \pm 7}{2}$

So the solutions are:

$x = \frac{13 + 7}{2} = 10 \quad \text{and} \quad x = \frac{13 - 7}{2} = 3$

These are the critical points at $x = 3$ and $x = 10$.

Step 3: Determine the nature of the critical points

To determine whether these are local minima or maxima, we can use the second derivative test.

Step 4: Find the second derivative

Differentiate $f'(x)$:

$f''(x) = 12x - 78$

Evaluate the second derivative at $x = 3$ and $x = 10$:

• For $x = 3$:

$f''(3) = 12(3) - 78 = 36 - 78 = -42$

Since $f''(3) < 0$, $x = 3$ is a local maximum.

• For $x = 10$:

$f''(10) = 12(10) - 78 = 120 - 78 = 42$

Since $f''(10) > 0$, $x = 10$ is a local minimum.

Step 5: Find the corresponding function values

Now substitute these $x$-values into the original function $f(x)$ to find the output values:

• For $x = 3$:

$f(3) = 2(3)^3 - 39(3)^2 + 180(3) + 9 = 2(27) - 39(9) + 540 + 9 = 54 - 351 + 540 + 9 = 252$

• For $x = 10$:

$f(10) = 2(10)^3 - 39(10)^2 + 180(10) + 9 = 2(1000) - 39(100) + 1800 + 9 = 2000 - 3900 + 1800 + 9 = -91$

Conclusion:

• The function has a local maximum at $x = 3$ with output value $252$.
• The function has a local minimum at $x = 10$ with output value $-91$.

Would you like more details or have any questions about the process?

Here are 5 related questions to explore:

1. How do you verify whether a critical point is a maximum or minimum using the first derivative test?
2. What does the second derivative test tell us about concavity?
3. How do you find inflection points for cubic functions?
4. What is the general form of a cubic function and how do its terms influence its shape?
5. Can you determine absolute extrema from critical points alone?

Tip: Always check the second derivative or use the first derivative test to classify critical points effectively!