Quadratic equation

2007-10-14 10:43 pm
1. Let f(x) = 4x^2-7x-k.
(a) Find the range of values of k for which f(x) >2 for all real values of x.
(b) Using the method of completing the square, show that f(x) attains its minimum value when x=7/8 for all values of k.

回答 (1)

2007-10-14 11:37 pm
✔ 最佳答案
(a) f(x) >2
f(x)-2 >0
4x^2 - 7x - (k+2) >0
Discriminant = (-7)^2 - 4(4)(-k-2) < 0
49 + 16(k+2) < 0
k+2 > 49/16
k>17/16

(b) f(x) = 4x^2 - 7x - k
f(x) = 4(x^2 - 7/4 x) - k
=4(x^2 - 7/4 x + (7/8)^2) - k -4(7/8)^2
= 4(x-7/8)^2 -k-49/16
>= -k-49/16 for all x (since (x-7/8)^2 >=0 for all x)
Therefore, f(x) attains its minimum when x = 7/8


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