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Question
The function f(x) = 2x3 – 3x2 – 12x + 4, has ______.
Options
Two points of local maximum
Two points of local minimum
One maxima and one minima
No maxima or minima
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Solution
The function f(x) = 2x3 – 3x2 – 12x + 4, has one maxima and one minima.
Explanation:
We have f(x) = 2x3 – 3x2 – 12x + 4
f'(x) = 6x2 – 6x – 12
For local maxima and local minima f'(x) = 0
∴ 6x2 – 6x – 12 = 0
⇒ x2 – x – 2 = 0
⇒ x2 – 2x + x – 2 = 0
⇒ x(x – 2) + 1(x – 2) = 0
⇒ (x + 1)(x – 2) = 0
x = –1, 2 are the points of local maxima and local minima
Now f'(x) = 12x – 6
`"f''"(x)_(x = -1)` = 12(–1) – 6
= – 12 – 6
= – 18 < 0, maxima
`"f''"(x)_(x = 2)` = 12(2) – 6
= 24 – 6
= 18 > 0 minima
So, the function is maximum at x = –1 and minimum at x = 2
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