To find the relative maximum and minimum of the function \( f(x) = x^3 + 3x^2 - 3x \), we first need to compute the first derivative and find the critical points:
- Find the first derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + 3x^2 - 3x) = 3x^2 + 6x - 3 \]
- Set the first derivative to zero to find critical points:
\[ 3x^2 + 6x - 3 = 0 \]
Dividing the equation by 3:
\[ x^2 + 2x - 1 = 0 \]
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]
So the critical points are:
\[ x_1 = -1 + \sqrt{2} \quad \text{and} \quad x_2 = -1 - \sqrt{2} \]
Calculating the approximate values:
\[ \sqrt{2} \approx 1.414 \]
Thus,
\[ x_1 \approx -1 + 1.414 \approx 0.414 \quad \text{(relative maximum candidate)} \] \[ x_2 \approx -1 - 1.414 \approx -2.414 \quad \text{(relative minimum candidate)} \]
- Evaluate the second derivative \( f''(x) \) to determine concavity:
\[ f''(x) = \frac{d}{dx}(3x^2 + 6x - 3) = 6x + 6 \]
Now evaluate \( f''(x) \) at the critical points:
- For \( x_1 \approx 0.414 \):
\[ f''(0.414) = 6(0.414) + 6 \approx 2.484 + 6 = 8.484 \quad (> 0) \Rightarrow \text{local minimum} \]
- For \( x_2 \approx -2.414 \):
\[ f''(-2.414) = 6(-2.414) + 6 \approx -14.484 + 6 = -8.484 \quad (< 0) \Rightarrow \text{local maximum} \]
- Find the corresponding function values at the critical points:
- For \( x_1 \approx 0.414 \):
\[ f(0.414) = (0.414)^3 + 3(0.414)^2 - 3(0.414) \approx 0.071 + 0.513 - 1.242 \approx -0.658 , \text{(rounded to -0.66)} \]
- For \( x_2 \approx -2.414 \):
\[ f(-2.414) = (-2.414)^3 + 3(-2.414)^2 - 3(-2.414) \approx -14.094 + 17.516 + 7.242 \approx 10.664 , \text{(rounded to 10.66)} \]
- Final Results:
Thus, the coordinates for the relative maximum are:
\(-2.41, 10.66\).
Thus, the correct answer is:
4) \((-2.41, 10.66)\)
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