Applications of Derivatives
Maxima & Minima of a Quartic
This one has three critical points and two separate minimum values — watch the walkthrough, then follow the full second‑derivative test below.
f(x) = 3x⁴ + 4x³ − 12x² + 12 — sketch of the three turning points
Example2 Worked Problem
Find the maximum and minimum values of $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.
Solution
Let $f(x)=3x^4+4x^3-12x^2+12$.
1 Find the critical values
$$f(x) = 3x^4 +4x^3 -12x^2 +12$$
$$f'(x) = 12x^3 + 12x^2 - 24x$$
$$f'(x) = 0 \implies 12x^3 + 12x^2 - 24x = 0$$
$$\implies 12x(x^2 + x - 2) = 0$$
$$\implies 12x(x + 2)(x - 1) = 0$$
$$\implies x = 0, \quad x = -2, \quad x = 1$$
Critical values are x = 0, x = −2 and x = 1
2 Apply the second‑derivative test
$$f'(x) = 12x^3 + 12x^2 - 24x$$
$$f''(x) = 36x^2 + 24x - 24$$
At x = −2
$$f''(-2) = 36(-2)^2 + 24(-2) - 24$$
$$= 144 - 48 - 24 = 72 > 0$$
f″ > 0 → Minimum point
At x = 0
$$f''(0) = 36(0)^2 + 24(0) - 24$$
$$= -24 < 0$$
f″ < 0 → Maximum point
At x = 1
$$f''(1) = 36(1)^2 + 24(1) - 24$$
$$= 36 + 24 - 24 = 36 > 0$$
f″ > 0 → Minimum point
3 Evaluate the maximum and minimum values
Minimum value 1 (at x = −2)
$$f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 12$$
$$f(-2) = -20$$
Maximum value (at x = 0)
$$f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 + 12$$
$$f(0) = 12$$
Minimum value 2 (at x = 1)
$$f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 12$$
$$f(1) = 7$$
Maximum Value
12
at x = 0
Minimum Value
−20
at x = −2
Minimum Value
7
at x = 1
✔✔
✦ ✦ ✦
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