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The maximum value of function f (x) = 2x ^ 3-2x ^ on [- 1,2]
f(x)=2x^3-2x^2
f'(x)=6x^2-4x
When f '(x) = 0, the solution is X1 = 0, X2 = 2 / 3
Both of them are in the domain of definition
f(-1)=-6-2=-8
f(2)=16-8=8
f(0)=0
f(2/3)=-8/9
So the maximum is f (2) = 8
Find the maximum value of the function f (x) = x + 2x + 3 on [T, t + 2]. Ask the great God for help
f(x)=(X+1)^2+2
Let f (x) satisfy the equation AF (x) + BF (x) = D of Cx + X, ABCD is constant, and the absolute value a is not equal to the absolute value B. find f (x) and prove the odd function of FX
Obviously, a + B is not equal to 0
That is, (a + b) f (x) = CX + D / X
f(x)=(cx+d/x)/(a+b)
That is, f (- x) = (- cx-d / x) / (a + b) = - f (x)
That is, f (x) is an odd function
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