To find the number of real solutions to the equation \( y = -x^3 - x^2 + 5x + 2 \), we can analyze the function \( f(x) = -x^3 - x^2 + 5x + 2 \) and determine its behavior.
-
Finding the derivative: We first calculate the derivative to understand the function's behavior: \[ f'(x) = -3x^2 - 2x + 5 \]
-
Finding critical points: Next, we set the derivative equal to zero to find the critical points: \[ -3x^2 - 2x + 5 = 0 \] To solve this quadratic equation, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -3 \), \( b = -2 \), and \( c = 5 \): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-3)(5)}}{2(-3)} \] \[ x = \frac{2 \pm \sqrt{4 + 60}}{-6} \] \[ x = \frac{2 \pm \sqrt{64}}{-6} \] \[ x = \frac{2 \pm 8}{-6} \] This gives us two critical points: \[ x = \frac{10}{-6} = -\frac{5}{3} \quad \text{and} \quad x = \frac{-6}{-6} = 1 \]
-
Evaluating the second derivative: To determine the nature of these critical points, we can evaluate the second derivative: \[ f''(x) = -6x - 2 \] Substituting the critical points into \( f''(x) \):
- At \( x = -\frac{5}{3} \): \[ f''\left(-\frac{5}{3}\right) = -6\left(-\frac{5}{3}\right) - 2 = 10 - 2 = 8 > 0 \quad \text{(local minimum)} \]
- At \( x = 1 \): \[ f''(1) = -6(1) - 2 = -6 - 2 = -8 < 0 \quad \text{(local maximum)} \]
-
Finding values at critical points: Now, we evaluate \( f(x) \) at these critical points:
-
At \( x = -\frac{5}{3} \): \[ f\left(-\frac{5}{3}\right) = -\left(-\frac{5}{3}\right)^3 - \left(-\frac{5}{3}\right)^2 + 5\left(-\frac{5}{3}\right) + 2 \] Calculating this: \[ = -\left(-\frac{125}{27}\right) - \frac{25}{9} - \frac{25}{3} + 2 \] (You can compute this to find the approximate value.)
-
At \( x = 1 \): \[ f(1) = -1 - 1 + 5 + 2 = 5 \]
-
-
Behavior as \( x \to \pm \infty \): As \( x \to -\infty \), \( f(x) \to +\infty \) because the leading term is negative. As \( x \to +\infty \), \( f(x) \to -\infty \).
By examining the local maximum and minimum, we can infer:
- \( f(x) \) starts high, dips down to a local minimum, rises to a local maximum, and then falls again. This suggests that there are three real roots since the graph crosses the x-axis three times.
Therefore, the equation \( y = -x^3 - x^2 + 5x + 2 \) has three real solutions.