Math
posted by Em on .
Given that f(x) = 2x^2  8x + 9, 0 ≤ x ≤ 3.
(a) Find the average value of f.
(b) Find c such that f(average) = f(c).

average value
= 1/(30) ∫(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