Worked Example · Applications of Derivatives
Maximum and Minimum Values via the Second Derivative Test
Find the maximum and minimum values of \( f(x) = 3x^4 + 4x^3 - 12x^2 + 12 \).
Step 1 · First Derivative
given\( f(x) = 3x^4 + 4x^3 - 12x^2 + 12 \)
diff\( f'(x) = 12x^3 + 12x^2 - 24x \)
Step 2 · Critical Points
set\( f'(x) = 0 \)
1\( 12x^3 + 12x^2 - 24x = 0 \)
2\( 12x\left(x^2 + x - 2\right) = 0 \)
3\( 12x(x + 2)(x - 1) = 0 \)
critical
points\( x = -2,\ 0,\ 1 \)
points\( x = -2,\ 0,\ 1 \)
Step 3 · Second Derivative Test
Differentiate \( f'(x) \) again: \( f''(x) = 36x^2 + 24x - 24 \). Evaluate at each critical point:
| x | \( f''(x) \) | Sign | Nature | \( f(x) \) |
|---|---|---|---|---|
| −2 | \( 36(4) + 24(-2) - 24 = 72 \) | + | Local minimum | −20 |
| 0 | \( 36(0) + 24(0) - 24 = -24 \) | − | Local maximum | 12 |
| 1 | \( 36(1) + 24(1) - 24 = 36 \) | + | Local minimum | 7 |
Conclusion
Maximum value
f(0) = 12
local maximum at \(x = 0\)
Minimum value
f(−2) = −20
global minimum at \(x = -2\)
Minimum value
f(1) = 7
local minimum at \(x = 1\)
Note: since \(f\) is defined on all of \(\mathbb{R}\) and \(x^4\) dominates for large \(|x|\), \(f(x) \to +\infty\) as \(x \to \pm\infty\), so \(f\) has no global maximum — only the local maximum at \(x=0\). The global minimum is the smaller of the two local minima, \(-20\).
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