JAMB - Mathematics (1992 - No. 34)
Obtain a maximum value of the function f(x) x\(^3\) - 12x + 11
-5
-2
2
27
Explanation
f(x) = x\(^3\) - 12x + 11
\(\frac{df(x)}{dx)}\) = 3x\(^2\) - 12 = 0
∴ 3x\(^2\) - 12 = 0 \(\to\) x\(^2\) = 4
x = \(\pm\)2, f(+2) = 8 - 24 + 11 = -15
= f(-2) = (-8) + 24 + 11
= 35 - 8 = 27
∴ maximum value = 27
\(\frac{df(x)}{dx)}\) = 3x\(^2\) - 12 = 0
∴ 3x\(^2\) - 12 = 0 \(\to\) x\(^2\) = 4
x = \(\pm\)2, f(+2) = 8 - 24 + 11 = -15
= f(-2) = (-8) + 24 + 11
= 35 - 8 = 27
∴ maximum value = 27
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