Urgently, ask the master to answer the function problem

Since ymax=2, ymin= -2, we can know a=2 Since in a period, x = 8, x = 5 8 when y obtains the maximum and minimum values respectively, let the period be t, then there is t 2 = 5 8- 8 = 2, so t =

And because t=2 w, w=2

In this way, the analytic formula can be reduced to y=2sin(2x+) and x=8 is substituted into :

2=2sin(4+) i.e. sin(4+)=1, i.e. 4+ =2k + 2(k z), =2k + 4(k z) is obtained from the range of: = 4

Therefore, the analytic formula is: y=2sin(2x+4).

Based on the maximum value of -2 and the minimum value of -2, we know that a=2

Because in a cycle, t=(5 rows 8-rows 8) 2=rows.

So w=2 row t=2

Bring in two points and w=2 to get sin(row 4+)=1, sin(5 rows 4+)=-1

It is obtained that = row 4 + 2k row, because 0 < "row 2, so k takes 0, so = row 4

So the analytic expression of the function is y=2sin(2x+row 4).

y=2sin(2x+pai/4)

The period is PAI, the amplitude is 2, and the displacement is 8

This can be compared with the following functions.

y=sinx: the period is 2pai, the amplitude is 1, and the displacement is 0

y=2sinx: the period is 2pai, the amplitude is 2, and the displacement is 0

y=2sin2x: the period is pai, the amplitude is 2, and the displacement is 0

y=2sin(2x+pai 4): the period is pai, the amplitude is 2, and the displacement is negative pai 8

The key is to plot the function on the xy axis.

f(x)=√3sin2x＋cos2x=2sin(2x＋π/6)。

The minimum positive period is , and the monotonic increase interval is [k 3, k 6], k z.

Over (0,1) yields c=1;f(x) is tangent to the straight line, indicating that there is only one intersection point, so substituting the straight line into the parabola gives ax 2+bx+1=1-2x, simplifying it to get ax 2+(b+2)x=0, and an intersection point gives b+2=0 to get b=-2;Finding the derivative once to get 2ax-2=0 gives that x is a pole at 1 a, and the opening upwards gives that the function is decreasing before 1 a, increasing when greater than and 1 a is at [1,3], the minimum is f(1 a)=1-1 a, the maximum is f(1)=a-1, f(3)=9a-5 points are discussed: when a-1>9a-5 gets a<, then g(x)=a-1-1+1 a=a+1 a-2=2 gives a no solution When a-1<9a-5 gets a>, then g(x)=9a-5-1+1 a=9a-6+1 a=2 can find a.

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According to the inequality formula x+1 x is greater than or equal to double the root number 1 x times x, i.e. 2. So the minimum value is 2, and when the right side of the equation contains x and the sub is 2, the obtained x value is the minimum value. I don't have any pen and paper around now, so if you don't understand, ask me again.