Given the function f (x) = e ^ x + X? - x, if the function y = | f (x) - t | - 3 has four zeros, then the value range of real number T is obtained Kneel down to ask the master to answer!!!!!!!

Given the function f (x) = e ^ x + X? - x, if the function y = | f (x) - t | - 3 has four zeros, then the value range of real number T is obtained Kneel down to ask the master to answer!!!!!!!

Find the extremum of F (x)
f'(x)=e^x+2x-1=0
X = 0, that is, the point (x1, Y1) = (0,1)
And f '' (x) = e ^ x + 2 > 0
Therefore, the point (x1, Y1) = (0,1) is the minimum point of F (x)
If you want y to have four zeros,
There must be | f (x1) - t- 3 > 0
f(x1)-t4

If the function f (x) = x? - 2 | x | - 3-A has four zeros, then the value range of real number a is

The function f (x) = x ^ 2-2-2 | x | - 3-A has four zeros, then the equation x ^ 2-2 | x | - 3-A = 0 has four roots, so equation x ^ 2-2 | x | = a + 3 has four roots, let Y1 = x ^ 2-2-2 | x |, y2 = a + 3 function Y1 = a + 3 function Y1 = x ^ 2-2 | x | image of y2 = x ^ 2-2 | x | x | pay attention to Y1 is a even function image image image of even function image about y axis symmetry about y axis symmetry graph know-1 | - 1 | - 1 (1)

Finally, the function f (x) = x? - 2 times the absolute value of x-3-a has four zeros. How to find the value of real number a

F (x) = x? - 2|x| - 3-A has four zeros
In other words, there are four different real number solutions for x? - 2 | x | - 3 = a
In other words, there are four intersections between the image of y = x? - 2|x | - 3 and the line y = a
When x ≥ 0,
y=x²-2|x|-3
=x²-2x-3
=(x-1) 2 - 4 draw the image,
When x < 0, the function image is symmetric with respect to the Y axis when x > 0,
If there are four intersections between the curve and the straight line, it is necessary to
-4

The minimum positive period of the function y = sin (2x - π / 2) sin2x

Sin (2x - π / 2) = - cos2x, y = - cos2xsin2x = - 1 / 2sin4x, so the minimum positive period is a two-way school

Given the function f (x) = sin (2x + π / 3), if f (x) = - 3 / 5, X belongs to (0, π / 2), find sin2x

Because x belongs to (0, π / 2),
Then, 2x + π / 3 belongs to (π / 3,4 π / 3)
By F (x) = - 3 / 5, sin (2x + π / 3) = - 3 / 5
Then we have π

The square of the function f (x) = sin (2x + π / 6) + sin2x Finding the maximum value of the minimum positive period of a function and the monotone increasing interval of the value set function of X when obtaining the maximum value The trouble will be the product of sum and difference. And the original formula is sin (2x + π / 6) + (the square of sin2x)

The sum difference product can be obtained as follows: F (x) = 2Sin (2x + π / 12) cos π / 12 = [(√ 2 + √ 6) / 2] sin (2x + π / 12). (1). T = 2 π / 2 = π. (2) f (x) max = (√ 2 + √ 6) / 2. In this case, x = k π + 5 π / 24. (k is an integer) (3) monotonically increasing interval is [K π - 7 π / 24, K π + 5 π / 24]. (k is an integer)

Given the function f (x) = sinxcosx + √ 3cos? X, find the minimum positive period of F (x)

This problem needs to use double angle formula and auxiliary angle formula
sinxcosx=1/2sin2x
√3cos²x=√3cos²x+√3/2-√3/2=√3（cos²x-1/2）+√3/2=√3/2cos2x+√3/2
∴f（x）=1/2sin2x+√3/2cos2x+√3/2=sin（2x+π/3）+√3/2
∴ T=2π/2=π

Find the period of the function f (x) = √ 3cos? X + sinxcosx Thank you = v=

f（x）=√3cos²x＋sinxcosx
=√3 (1+cos2x)/2+1/2 sin2x
=√3/2 cos2x+1/2 sin2x+√3/2
=sinπ/3 cos2x+cosπ/3 sin2x+√3/2
=sin(π/3+2x)+√3/2
So period T = 2 π / 2 = π
Answer: the period is π

F (x) = 2 √ 3cos? X-2sinxcosx - √ 3 (1) find the minimum positive period and minimum value of the function (2) find the monotone increasing interval of the function f (x) Find it quickly

f(x)=2√3(1+cos2x)/2-sin2x-√3
=-(sin2x-√3cos2x)
=-2sin(2x-π/3)
So t = 2 π / 2 = π
The minimum value is - 2
The increase of sin (2x - π / 3) decreases
So 2K π + π / 2

Find the maximum and minimum positive period of the function y = 2cos? X + sin2x-1

According to the formula of double angle: cos2x = 2cos? X-1
The result is: y = 2cos? X + sin2x-1
=cos2x+sin2x
=√2sin(2x+π/4)
So, the minimum positive period is: 2 π / 2 = π
When 2x + π / 4 = 2K π + π / 2, that is, x = k π + π / 8, the maximum value of Y is: √ 2
When 2x + π / 4 = 2K π - π / 2, that is, x = k π - 3 π / 8, the minimum value of Y is: - √ 2