When t ≤ x ≤ T + 1, find the function y = 1 2x2-x-5 2 (where t is a constant)

When t ≤ x ≤ T + 1, find the function y = 1 2x2-x-5 2 (where t is a constant)

∵ the symmetry axis equation of the image with function y = 12x2-x-52 = 12 (x-1) 2-3 is x = 1. When t + 1 < 1, the function is a subtractive function on [T, t + 1], so the maximum value of the function is f (T) = 12t2-t-52 and the minimum value is f (T + 1) = 12t2-3. When t ≤ 1 < T + 12, the maximum value of the function is f (T + 1) = 12t2-3

In interval[ ½, 2] On, function f (x) = x ²+ PX + Q and G (x) = 2x + 2 / X obtain the same minimum value at the same point, then what is the value of Q

Because in the interval[ ½, 2] Inside
g(x)=2x+2／x≥4
At this time, 2x = 2 / x, that is, x = 1
For function f (x) = x ²+ PX + Q and G (x) = 2x + 2 / X obtain the same minimum value at the same point
So when x = - P / 2 = 1, that is, P = - 2
f(x)min=4q-p ²/ 4=4 q=5

If x > 1 is known, the function f (x) = (x) ²- The minimum value of 2x + 2) / (x-1) is

F(X)=(x ²- 2x+2)/(x-1)
=(X-1)^2+1/(X-1)
=X-1+1/(X-1)
》2 if and only if x = 2=

Known function f (x) = (x) ²+ 2X + a) / x, X ∈ [1, + ∞). (1) when a = 4, find the minimum value of F (x)

∵ x > 0 ∵ f (x) = (x ^ 2 + 2x + a) / x = x + 2 + A / X ≥ 2 √ a + 2. When a = 4, f (x) ≥ 2 √ 4 + 2 = 6, then x = 2. And f (x) = x + 4 / x + 2, you can find the derivative f '(x) = 1-4 / X ², On [1,2], f '(x) ≤ 0, so it is a subtractive function
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Given the function FX = sin (5 / 4 pie - x) - cos (4 / 4 pie x), find the monotonic decreasing interval of FX Given the function FX = sin (5 / 4 pie - x) - cos (4 / 4 pie + x) 1, find the monotonic increasing interval of FX, and 2 know cos（ α-β)= 3 / 5, cos（ α+β)- 3 / 5 ＜ α＜β＜ Two thirds of the pie, ask F β

f(x)=sin(5π/4-x)-cos(π/4+x)
=sin(π+π/4-x)-cos(π/4+x)
=-sin(π/4-x)-cos(π/4-x)
=-√2(√2/2sin(π/4-x)+√2/2cos(π/4-x))
=-√2sin(π/4-x+π/4)
=-√2sin(π/2-x)
=-√2cosx
∵ f (x) decreasing
‡ cosx increment
∴2kπ-π<=x<=2kπ
0＜ α＜ b＜π/2
-π/2<-b<0
-π/2cos(a-b)=3/5
sin(a-b)=-√(1-cos^2(a-b))=-4/5
0cos(a+b)=-3/5
sin(a+b)=√(1-cos^2(a+b))=4/5
cos2b=cos(a+b-(a-b))
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=-3/5*3/5 +4/5*(-4/5)
=-9/25-16/25
=-1
=2cos^2b-1
cosb=0
B = π / 2 (the meaning of the question doesn't say B < = π / 2, it's strange)
f(b)=-√2cosb=0

Given the function f (x) = 3x alnx (a belongs to R), the monotone interval and extreme point of function f (x) are discussed If the function f (x) has an extreme point x0, the slope of the straight line between the demerit point a (x0, f (x0)) and the origin is K. is there a so that k = 3-A? If it exists, find the value of a; If not, please explain the reason

f(x)=3x-alnx f'(x)=3-a/x
F '(x) = 3-A / x0 = 0 at extreme point x0, so a = 3x0
k=f(x0)/x0=(3x0-alnx0)/x0=(3x0-3x0lnx0)/x0=3-3lnx0
If k = 3-A, 3-3lnx0 = 3-3x0
lnx0=x0
g(x)=x-lnx g'(x)=1-1/x
At 00 g (x), x > 1 is a single increasing function, and G (1) = 0, so when x > 1, G (x) > G (1) = 1 > 0
Therefore, there is no X so that G (x) = 0, so there is no a so that k = 3-a