Consider the function 𝑓: ℝ → ℝ where ℝ is the set of all real numbers. \(f(x)…

For a function that is differentiable everywhere, local extrema occur only at critical points — points where the first derivative equals zero. The second-derivative test then classifies each critical point: a positive second derivative there means a local minimum, and a negative second derivative there means a local maximum.

1. Differentiate f(x) = x4/4 − 2x3/3 − 3x2/2 + 1 to get f′(x) = x3 − 2x2 − 3x.

2. Factor: f′(x) = x(x2 − 2x − 3) = x(x − 3)(x + 1). Setting f′(x) = 0 gives the critical points x = −1, 0, and 3.

3. Differentiate again: f″(x) = 3x2 − 4x − 3.

4. Evaluate f″ at each critical point — f″(−1) = 3(1) − 4(−1) − 3 = 4 (positive, so a local minimum); f″(0) = 3(0) − 4(0) − 3 = −3 (negative, so a local maximum); f″(3) = 3(9) − 4(3) − 3 = 12 (positive, so a local minimum).

Cross-check with the sign of f′(x) = x(x − 3)(x + 1) across the intervals its roots create: negative for x < −1, positive on (−1, 0), negative on (0, 3), and positive for x > 3. f′ flips from negative to positive at x = −1 (confirming a minimum there), from positive to negative at x = 0 (confirming a maximum there), and from negative to positive at x = 3 (confirming a minimum there) — matching the second-derivative test exactly.

So x = 0 is a local maximum and x = 3 is a local minimum of f, while x = −1 is a local minimum (not a maximum) and x = 0 is not a local minimum.