57. \( f(x)=x^{3}+5 x^{2}-8 x+2 \)

To analyze the function \( f(x) = x^3 + 5x^2 - 8x + 2 \), we can perform several steps, including finding its derivative, critical points, and determining its behavior. ### Step 1: Find the derivative The first derivative of the function \( f(x) \) is given by: \[ f'(x) = \frac{d}{dx}(x^3 + 5x^2 - 8x + 2) = 3x^2 + 10x - 8 \] ### Step 2: Find critical points To find the critical points, we set the derivative equal to zero: \[ 3x^2 + 10x - 8 = 0 \] We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 10 \), and \( c = -8 \): \[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} \] \[ x = \frac{-10 \pm \sqrt{100 + 96}}{6} \] \[ x = \frac{-10 \pm \sqrt{196}}{6} \] \[ x = \frac{-10 \pm 14}{6} \] This gives us two solutions: 1. \( x = \frac{4}{6} = \frac{2}{3} \) 2. \( x = \frac{-24}{6} = -4 \) ### Step 3: Determine the nature of critical points To determine whether these critical points are local maxima, minima, or points of inflection, we can use the second derivative test. The second derivative is: \[ f''(x) = \frac{d}{dx}(3x^2 + 10x - 8) = 6x + 10 \] Now we evaluate the second derivative at the critical points: 1. For \( x = \frac{2}{3} \): \[ f''\left(\frac{2}{3}\right) = 6\left(\frac{2}{3}\right) + 10 = 4 + 10 = 14 \quad (\text{positive, so local minimum}) \] 2. For \( x = -4 \): \[ f''(-4) = 6(-4) + 10 = -24 + 10 = -14 \quad (\text{negative, so local maximum}) \] ### Step 4: Evaluate the function at critical points Now we can find the values of the function at these critical points: 1. \( f\left(\frac{2}{3}\right) \): \[ f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 + 5\left(\frac{2}{3}\right)^2 - 8\left(\frac{2}{3}\right) + 2 \] \[ = \frac{8}{27} + 5 \cdot \frac{4}{9} - \frac{16}{3} + 2 \] \[ = \frac{8}{27} + \frac{20}{9} - \frac{144}{27} + \frac{54}{27} \] \[ = \frac{8 + 60 - 144 + 54}{27} = \frac{-22}{27} \] 2. \( f(-4) \): \[ f(-4) = (-4)^3 + 5(-4)^2 - 8(-4) + 2 \] \[ = -64 + 80 + 32 + 2 = 50 \] ### Summary of Results - Critical points: - \( x = \frac{2}{3} \) is a local minimum with \( f\left(\frac{2}{3}\right) = -\frac{22}{27} \). - \( x = -4 \) is a local maximum with \( f(-4) = 50 \). ### Conclusion The function \( f(x) = x^3 + 5x^2 - 8x + 2 \) has a local maximum at \( (-4, 50) \) and a local minimum at \( \left(\frac{2}{3}, -\frac{22}{27}\right) \).