標題:
Quadratic equation
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.
最佳解答:
(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
其他解答:
Quadratic equation
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發問: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.
最佳解答:
(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
其他解答:
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