Posted by Em on Sunday, January 27, 2013 at 3:16pm.
average value
= 1/(3-0) ∫(2x^2 - 8x + 9) dx from x = 0 to x = 3
= (1/3) [(2/3)x^3 - 4x^2 + 9x[ from 0 to 3
= (1/3) (18 - 36 + 27 - 0 )
= 3
b)
f(c) = f(3)
2c^2 - 8c + 9 = 18 - 16 + 9
2c^2 - 8c -2 = 0
c^2 - 4c - 1 = 0
c = (4 ± √20)/2
= 2 ± √5
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