allery$f(x) = 2x^3 - 3x^2 - 12x + 5$. The **local maximum value** occurs at:
A$x = -1$ (local max value 12)
B$x = 2$ (local min)
C$x = 0$
D$x = 1$
Answer & Solution
Correct answer: A. $x = -1$ (local max value 12)
We have $f'(x) = 6x^2 - 6x - 12$.
That is $f'(x) = 6(x-2)(x+1)$.
So $f'(x) = 0$.
⇒ $x = -1, 2$. $f''(x) = 12x - 6$.
At $x = -1$: $f''(-1) = -18 < 0$ ⇒ local max.
At $x = 2$: $f''(2) = 18 > 0$
⇒ local min. $f(-1) = -2 - 3 + 12 + 5 = 12$.
Keep practising Unit 4-5: Applications of Derivatives & Curve Sketching
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