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.
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Compute the derivative of \( f \): \[ f'(x) = 3x^2 + 6x - 3 \]
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Set the derivative equal to zero to find critical points: \[ 3x^2 + 6x - 3 = 0 \] Dividing the entire equation by 3 gives: \[ x^2 + 2x - 1 = 0 \]
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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: \[ x_1 = -1 + \sqrt{2} \quad \text{and} \quad x_2 = -1 - \sqrt{2} \]
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Calculate the numerical values: \[ x_1 \approx -1 + 1.414 \approx 0.414 \quad (\text{about } 0.41 \text{ when rounded}) \] \[ x_2 \approx -1 - 1.414 \approx -2.414 \quad (\text{about } -2.41 \text{ when rounded}) \]
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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.
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Compute the second derivative: \[ f''(x) = 6x + 6 \]
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Evaluate the second derivative at the critical points:
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For \( x_1 = -1 + \sqrt{2} \): \[ f''(-1 + \sqrt{2}) = 6(-1 + \sqrt{2}) + 6 = 6\sqrt{2} \approx 8.49 > 0 \quad (\text{indicating a local minima}) \]
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For \( x_2 = -1 - \sqrt{2} \): \[ f''(-1 - \sqrt{2}) = 6(-1 - \sqrt{2}) + 6 = -6\sqrt{2} < 0 \quad (\text{indicating a local maxima}) \]
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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: \[ f(-2.41) \text{ and } f(0.41) \]
- 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:
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The coordinates of the relative maximum are \( (-2.41, 10.66) \) which corresponds to:
- Option 4) \((-2.41, 10.66)\)
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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: \((0.41, -0.66)\)
1 answer