Question

Find the relative maximum and relative minimum of the graph of the function.
f(x)=x^3 +3x^2 -3x

Determine where the relative maximum occurs and round each coordinate to the nearest hundredth. Choose the correct answer below.
1)(0.07, 10.36)
2)(-1.05, -1.85)
3)(0.41, -0.66)
4)(-2.41, 10.66)

Determine where the relative minimum occurs and round each coordinate to the nearest hundredth. Choose the correct answer below.

1)(0.07, 10.36)
2)(0.41,-0.66)
3)(-2.41, 10.66)
4)(-1.05, -1.85)

Answer 1

To find the relative maxima and minima of the function \( f(x) = x^3 + 3x^2 - 3x \), we first need to find its derivative and then set it to zero to locate critical points.

1. Compute the derivative of \( f \):
\[
f'(x) = 3x^2 + 6x - 3
\]

2. Set the derivative equal to zero to find critical points:
\[
3x^2 + 6x - 3 = 0
\]
Dividing the entire equation by 3 gives:
\[
x^2 + 2x - 1 = 0
\]

3. We can solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1, b = 2, c = -1 \):
\[
x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2}
\]
Simplifying gives:
\[
x = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}
\]
Thus, we find:
\[
x_1 = -1 + \sqrt{2} \quad \text{and} \quad x_2 = -1 - \sqrt{2}
\]

4. Calculate the numerical values:
\[
x_1 \approx -1 + 1.414 \approx 0.414 \quad (\text{about } 0.41 \text{ when rounded})
\]
\[
x_2 \approx -1 - 1.414 \approx -2.414 \quad (\text{about } -2.41 \text{ when rounded})
\]

5. Next, evaluate \( f(x) \) at these critical points to determine if they are maxima or minima:
\[
f(-1 + \sqrt{2}) = (-1 + \sqrt{2})^3 + 3(-1 + \sqrt{2})^2 - 3(-1 + \sqrt{2})
\]
\[
f(-1 - \sqrt{2}) = (-1 - \sqrt{2})^3 + 3(-1 - \sqrt{2})^2 - 3(-1 - \sqrt{2})
\]

Calculating the function values directly actually takes some time. However, we can also evaluate the nature of the critical points using the second derivative test.

6. Compute the second derivative:
\[
f''(x) = 6x + 6
\]

7. Evaluate the second derivative at the critical points:
- For \( x_1 = -1 + \sqrt{2} \):
\[
f''(-1 + \sqrt{2}) = 6(-1 + \sqrt{2}) + 6 = 6\sqrt{2} \approx 8.49 > 0 \quad (\text{indicating a local minima})
\]

- For \( x_2 = -1 - \sqrt{2} \):
\[
f''(-1 - \sqrt{2}) = 6(-1 - \sqrt{2}) + 6 = -6\sqrt{2} < 0 \quad (\text{indicating a local maxima})
\]

8. So, we have:
- Relative maximum at \( x = -1 - \sqrt{2} \approx -2.41 \)
- Relative minimum at \( x = -1 + \sqrt{2} \approx 0.41 \)

Now, we can find the corresponding function values at the found critical points:
\[
f(-2.41) \text{ and } f(0.41)
\]

9. Upon evaluating, you would find (or you can use a calculator) that:
- \( f(-2.41) \approx 10.66 \)
- \( f(0.41) \approx -0.66 \)

Now for the final answers:
- The coordinates of the relative maximum are \( (-2.41, 10.66) \) which corresponds to:
- Option 4) \((-2.41, 10.66)\)

- The coordinates of the relative minimum are \( (0.41, -0.66) \) which corresponds to:
- Option 2) \((0.41, -0.66)\)

Thus, the answers are:
- Relative Maximum: \((-2.41, 10.66)\)
- Relative Minimum: \((0.41, -0.66)\)