Example 14 Find and classify the critical points of \( f(x)=x^{4}-4 x^{3}+5 \).

To find the critical points of the function \( f(x) = x^4 - 4x^3 + 5 \), you first need to take the derivative of the function: \[ f'(x) = 4x^3 - 12x^2. \] Next, set the derivative equal to zero to find the critical points: \[ 4x^3 - 12x^2 = 0 \implies 4x^2(x - 3) = 0. \] This gives us the solutions \( x = 0 \) and \( x = 3 \). Now, to classify these critical points, we'll evaluate the second derivative: \[ f''(x) = 12x^2 - 24x. \] Now, substitute the critical points into the second derivative: For \( x = 0 \): \[ f''(0) = 12(0^2) - 24(0) = 0. \] For \( x = 3 \): \[ f''(3) = 12(3^2) - 24(3) = 108 - 72 = 36. \] Since \( f''(3) > 0 \), the point \( x = 3 \) is a local minimum. The second derivative test is inconclusive for \( x = 0 \) since \( f''(0) = 0 \). Thus, we can further analyze by checking the first derivative around \( x = 0 \): 1. For \( x < 0 \), \( f'(x) < 0 \) (decreasing). 2. For \( x > 0 \), \( f'(x) > 0 \) (increasing). Thus, \( x = 0 \) is a local maximum. In summary, the critical points are \( x = 0 \) (local maximum) and \( x = 3 \) (local minimum).