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If f (x + 2) = x-x + 1, then f (x) is equal to
Let t = x + 2, then x = t-2f (T) = (T-2) &# 178; - (T-2) + 1 = T & # 178; - 4T + 4-T + 2 + 1 = T & # 178; - 5T + 7, that is, f (x) = x & # 178; - 5x + 7
What is the maximum value of function f (x) = x ^ 3-x ^ 2-x + 1 when x belongs to [- 1,1]?
Take the derivative of F (x) and make the derivative equal to 0, that is: F '(x) = 3 * x ^ 2 + 2 * X-1 = 0, then x = - 1 / 3 x = 1
That is to say, in (- ∞, - 1 / 3) f (x) monotone increase, in (- 1 / 3,1) f (x) monotone decrease, in (1, ∞) f (x) monotone increase
Therefore, when [- 1,1] f (x) is the maximum, x = - 1 / 3, then f (- 1 / 3) = 32 / 27
If f (x-1 / x) = x + 1 / x, then f (3) is equal to?
f(x-1/x)=√[(x-1/x)^2+4]
f(3)=√(3^2+4))=√13
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