數學的值域要怎麼找?

Question

設函數f:R→R,f(x)=2sin^2 x+ cos x - 1,試求f(x)之值域?

Answer 1

f(x) = 2sin² x + cos x - 1
= 2(1 - cos² x ) + cos x - 1
= 2 - 2cos² x + cos x - 1
= -2cos² x + cos x + 1
= -2(cos² x - (1/2)cos x - 1/2)
= -2(cos² x - (1/2)cos x + (1/4)² - (1/4)² - 1/2)　(國中配方法)
= -2((cos x - 1/4)²- 1/16 - 1/2)
= -2((cos x - 1/4)² - 9/16)
= -2(cos x - 1/4)² + 9/8　(這裡已可看出最大 9/8)
又 -1 ≤ cos x ≤ 1
-1 - 1/4 ≤ cos x - 1/4 ≤ 1 - 1/4
-5/4 ≤ cos x - 1/4 ≤ 3/4
0 ≤ (cos x - 1/4)² ≤ 25/16
-2‧25/16 ≤ -2(cos x - 1/4)² ≤ 0
-25/8 ≤ -2(cos x - 1/4)² ≤ 0
-25/8 + 9/8 ≤ -2(cos x - 1/4)² + 9/8 ≤ 9/8
-16/8 ≤ -2(cos x - 1/4)² + 9/8 ≤ 9/8
-2 ≤ -2(cos x - 1/4)² + 9/8 ≤ 9/8
-2 ≤ f(x) ≤ 9/8　(值域)