To determine how many real solutions the equation \( y = -x^3 - x^2 + 5x + 2 \) has, we need to analyze the cubic function \( f(x) = -x^3 - x^2 + 5x + 2 \).
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Finding the Derivative: First, we find the derivative to analyze its critical points: \[ f'(x) = -3x^2 - 2x + 5 \]
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Setting the Derivative to Zero: Now we set the derivative equal to zero to find critical points: \[ -3x^2 - 2x + 5 = 0 \] Multiplying by \(-1\) gives: \[ 3x^2 + 2x - 5 = 0 \]
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Using the Quadratic Formula: We can use the quadratic formula to solve for \(x\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \] \[ = \frac{-2 \pm \sqrt{4 + 60}}{6} = \frac{-2 \pm \sqrt{64}}{6} = \frac{-2 \pm 8}{6} \] Thus, we have two solutions: \[ x_1 = \frac{6}{6} = 1, \quad x_2 = \frac{-10}{6} = -\frac{5}{3} \]
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Analyzing the Critical Points: Now we evaluate the function \( f(x) \) at these critical points to determine the local maxima and minima.
- \( f(1) = -1^3 - 1^2 + 5(1) + 2 = -1 - 1 + 5 + 2 = 5 \).
- \( f\left(-\frac{5}{3}\right) \) is harder to calculate directly, but we only need to know that there are critical points.
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Behavior as \( x \to \infty \) and \( x \to -\infty \):
- As \( x \to \infty \), \( f(x) \to -\infty \) (since the leading term is \(-x^3\)).
- As \( x \to -\infty \), \( f(x) \to \infty \).
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Analyzing the Values at Critical Points: The function changes from positive to negative based on our evaluations, which indicates points where the function crosses the x-axis.
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Conclusion Using Intermediate Value Theorem: Since \( f(x) \to \infty \) as \( x \to -\infty \) and \( f(x) \to -\infty \) as \( x \to \infty \), and since the function has two critical points:
- There will be at least one root between \( -\infty \) and the highest point (a maximum) near \( f(1) = 5 \).
- And there will be at least one root between \( 1 \) (where it is positive) and \( -\frac{5}{3} \) before it goes negative.
Overall, there are three real solutions to the equation \( y = -x^3 - x^2 + 5x + 2 \).