Identify and classify the stationary points of f using the second derivative test, where f is the function given below
f(x) = x4- 4x3+ 4x2
Given the function above, we need to find its stationary points. Its stationary points occur either at the minima and maxima of the function or at the endpoints. Noticing that the domain of the function is from positive infinity to negative infinity, we proceed to find the maxima and minima of the graph by finding the points at which the tangent like to the graph has slope 0. We will do this by taking the first derivative and equating to zero:
f'(x) = 4x3 - 12x2 + 8x = 0
Simplifying: 0 = 4x(x2 - 3x + 2) = 4x (x - 1)(x - 2)
Hence, the function's stationary points occur at: x = 0, x = 1 and x = 2 Since all points belong to the domain, we proceed to classify them using the second derivative test:Finding the second derivative of the function:f''(x) = 12x2 - 24x + 8 = 4 (3x2 - 6x + 2) Then, to classify the stationary points we need to find the second derivative of the function at each of the points: For x = 0: f''(0) = 8 Since 8 > 0, we conclude that x = 0 is a local minimumFor x = 1: f''(1) = - 4 Since - 4 < 0, we conclude that x = 1 is a local maximum For x = 2: f''(2) = 8 Since 8 > 0, we conclude that x = 2 is a local minimum
