Question

Given \(f(x) = 3x^4 + 4x^3 - 12x^2 + 6\), the minimum value of \(f(x)\) is ______.

• A. \(-6\)
• B. \(-36\)
• C. \(-26\)
• D. \(-46\)

Hint:

Differentiate, factor f'(x) = 12x(x+2)(x-1) to get the critical points, use the second derivative to identify which are minima, then compare f(x) at those points.

Answer 1

The Correct Option is C

Step 1: Find the critical points by differentiating. \[f'(x) = 12x^3 + 12x^2 - 24x\] Factor out common terms: \[f'(x) = 12x(x^2 + x - 2) = 12x(x+2)(x-1)\] Setting \(f'(x) = 0\) gives \(x = 0, -2, 1\).

Step 2: Classify each critical point using the second derivative. \[f''(x) = 36x^2 + 24x - 24\] At \(x=0\): \(f''(0) = -24 < 0\), so this is a local maximum.
At \(x=-2\): \(f''(-2) = 36(4) + 24(-2) - 24 = 144 - 48 - 24 = 72 > 0\), so this is a local minimum.
At \(x=1\): \(f''(1) = 36 + 24 - 24 = 36 > 0\), so this is also a local minimum.

Step 3: Evaluate f(x) at both local minima. \[f(-2) = 3(16) + 4(-8) - 12(4) + 6 = 48 - 32 - 48 + 6 = -26\] \[f(1) = 3(1) + 4(1) - 12(1) + 6 = 3 + 4 - 12 + 6 = 1\]

Step 4: Confirm which is the global minimum.
Since the leading term \(3x^4\) has a positive coefficient and even degree, \(f(x) \to +\infty\) as \(x \to \pm\infty\), so the smallest value of f(x) anywhere must occur at one of the local minima found above. Comparing \(f(-2) = -26\) and \(f(1) = 1\), the global minimum is \(-26\), which is option (C). Options (A), (B), and (D) do not correspond to the correct value of f at either genuine local-minimum point.