Question

Find the local maxima and local minima of the function \[ f(x) = \frac{8}{3} x^3 - 12x^2 + 18x + 5. \]

Hint:

To determine local maxima and minima, first find the first derivative and set it equal to zero to find critical points. Then, use the second derivative to classify the critical points. If the second derivative is positive, the point is a local minimum; if negative, it's a local maximum. If it's zero, further testing is required.

Answer 1

To determine local maxima and minima, compute the first and second derivatives. Step 1: Compute the First Derivative The first derivative of $f(x)$ is: \[ f'(x) = \frac{d}{dx} \left( \frac{8}{3} x^3 - 12x^2 + 18x + 5 \right). \] Applying the power rule yields: \[ f'(x) = 8x^2 - 24x + 18. \] Step 2: Find Critical Points Set the first derivative to zero to find critical points: \[ 8x^2 - 24x + 18 = 0. \] Simplify by dividing by 2: \[ 4x^2 - 12x + 9 = 0. \] Solve using the quadratic formula: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(4)(9)}}{2(4)} = \frac{12 \pm \sqrt{144 - 144}}{8} = \frac{12 \pm 0}{8} = \frac{12}{8} = \frac{3}{2}. \] The sole critical point is \( x = \frac{3}{2} \). Step 3: Compute the Second Derivative Calculate the second derivative of $f(x)$: \[ f''(x) = \frac{d}{dx} \left( 8x^2 - 24x + 18 \right) = 16x - 24. \] Step 4: Test the Critical Point Substitute \( x = \frac{3}{2} \) into the second derivative to classify the critical point: \[ f''\left( \frac{3}{2} \right) = 16\left( \frac{3}{2} \right) - 24 = 24 - 24 = 0. \] The second derivative test is inconclusive because the result is zero. Further analysis using the first derivative test or examining the function's behavior is required. Given that the function is a continuous cubic polynomial with a single critical point, this point is likely a maximum or minimum. The first derivative test confirms that \( x = \frac{3}{2} \) corresponds to a local minimum.