How to calculate (- x + 2x & # 178; + 5) + (4x & # 178; - 3-6x)? (- x + 2x & # 178; + 5) + (4x & # 178; - 3-6x) how to calculate? What's more, how to confirm the sign? The teacher spoke too fast and didn't hear it clearly

How to calculate (- x + 2x & # 178; + 5) + (4x & # 178; - 3-6x)? (- x + 2x & # 178; + 5) + (4x & # 178; - 3-6x) how to calculate? What's more, how to confirm the sign? The teacher spoke too fast and didn't hear it clearly

(-x+2x²+5）+（4x²-3-6x）
=-x+2x²+5+4x²-3-6x
=6x²-7x+2
The maximum and minimum of function y = x / X & # 178; + 1!
Using function to find
The first one is the first one
Let t = x / (X & # 178; + 1)
t(x²+1) = x
tx² - x +t = 0
From Δ > = 0, we get (- 1) &# 178; - 4T & # 178; > = 0
The solution is t-178;
base number
And he > o
That's it
The minimum range of Y is not equal to 1? I don't think so
Given the function f (x) = - 3sin ^ 2x-4cosx + 2 (1), find the value of F (PI / 3) and (2) find the maximum and minimum value of F (x)
(1) The equation: F (x) (x) = -3sin \35\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\and
f(π/3)=-9/4-2+2=-9/4
f(x)=-3(1-cos²x)-4cosx+2
=-3cos²x-4cosx+5
=-3(cosx+2/3)²+19/3
So cosx = - 2 / 3, maximum = 19 / 3
Cosx = 1, minimum = - 2
1.f(π /3)=-3(3/4)-4*1/2+2=-9/4-2+2=-9/4
2. Let t = cosx, f (x) = - 3 (1-T ^ 2) - 4T + 2 = 3T ^ 2-4t-1 = 3 (T-2 / 3) ^ 2-7 / 3
When t = 2 / 3, the minimum value is - 7 / 3
When t = - 1, the maximum value is 6
f(x)=-3sin^2x-4cosx+2
=-3(1-cos^2x)-4cosx+2
=-3+3cos^2x-4cosx+2
=3cos^2x-4cosx-1
=3(cosx-2/3)^2-1-4/3
=3(cosx-2/3)^2-7/3
F (Pie / 3) = 3 (1 / 2-2 / 3) ^ 2-7 / 3
=3(-1/6)^2-7/3
=1/12-28/12
=-27/12
=-9/4
f(x)=3(cosx-2/3)^2-7/3
When cosx = 2 / 3, the minimum value is - 7 / 3
When cosx = - 1, the maximum value is 6
Finding the maximum and minimum of F (x) with known function f (x) = - 3sin ^ 2-4cosx + 2
F (x) = - 3 (1-cos ^ 2) - 4cosx + 2, let t = cosx (- 1)
f(x)=-3sinx^2-4cosx+2
=-3(1-cos²x)-4cosx+2
=3cos²x -4cosx-1
=3(cosx-2/3)²-7/3
cosx∈[-1,1],
=> f(x)max=3(-1-2/3)²-7/3=6,
f(x)min=-7/3
f(x)=-3sin^2-4cosx+2=3cosx^2-4cosx-1=3(cosx-2/3)^2-7/3
The minimum value is 7 / 7
When the maximum value is cosx = - 1, f (x) = 6
f(x)=-3(1-cos²x)-4cosx+2
=3cos²x-4cosx-1
=3(cosx-2/3)²-7/3
So cosx = 2 / 3, minimum = - 7 / 3
Cosx = - 1, maximum = 6