Determine the location of each local extremum of the function. \( f(x)=\frac{4}{3} x^{3}-2 x^{2}-80 x+5 \)

To find the local extremum of the function \( f(x) = \frac{4}{3} x^{3} - 2 x^{2} - 80 x + 5 \), we start by calculating the derivative \( f'(x) \): \[ f'(x) = 4x^{2} - 4x - 80. \] Next, we set the derivative equal to zero to find critical points: \[ 4x^{2} - 4x - 80 = 0. \] Dividing the entire equation by 4 simplifies it to: \[ x^{2} - x - 20 = 0. \] Now, we can factor the quadratic: \[ (x - 5)(x + 4) = 0. \] Solving this gives us the critical points \( x = 5 \) and \( x = -4 \). To determine whether these points are local maxima or minima, we can use the second derivative test. The second derivative is: \[ f''(x) = 8x - 4. \] Now we'll evaluate the second derivative at each critical point: 1. For \( x = 5 \): \[ f''(5) = 8(5) - 4 = 40 - 4 = 36 > 0. \] Since \( f''(5) > 0 \), there is a local minimum at \( x = 5 \). 2. For \( x = -4 \): \[ f''(-4) = 8(-4) - 4 = -32 - 4 = -36 < 0. \] Since \( f''(-4) < 0 \), there is a local maximum at \( x = -4 \). Now we can find the corresponding \( y \)-values for these \( x \)-values by substituting back into the original function: 1. For \( x = 5 \): \[ f(5) = \frac{4}{3}(5^{3}) - 2(5^{2}) - 80(5) + 5 = \frac{4}{3}(125) - 50 - 400 + 5 = \frac{500}{3} - 50 - 400 + 5 = \frac{500 - 150 - 1200 + 15}{3} = \frac{-835}{3} \approx -278.33. \] 2. For \( x = -4 \): \[ f(-4) = \frac{4}{3}(-4^{3}) - 2(-4^{2}) - 80(-4) + 5 = \frac{4}{3}(-64) - 32 + 320 + 5 = -\frac{256}{3} - 32 + 320 + 5 = -\frac{256}{3} - \frac{96}{3} + \frac{960}{3} + \frac{15}{3} = \frac{623}{3} \approx 207.67. \] Thus, the local extrema are: - Local maximum at \( (-4, \frac{623}{3}) \) or approximately \( (-4, 207.67) \). - Local minimum at \( (5, -\frac{835}{3}) \) or approximately \( (5, -278.33) \).