The range of y = 1 / (x power of 3 + 1) is

The range of y = 1 / (x power of 3 + 1) is

Power X of 3 + 1 > 0 + 1 = 1
therefore
y＜1/1=1
Y > 0
therefore
The value range is (0,1)

Find the X + 1 power + 3 of y = 4 to the power of X-2, and X belongs to the value range of (negative infinity, 1]

Let a = 2 ^ x
Then 4 ^ x = a ²
2^(x+1)=2a
x

Y = 2 to the power of x minus 1 / 1 to find the value range,

y=1/(2^x-1)
Move to y × 2^x - y = 1
2^x=(1+y)/y
Take logarithm on both sides
x=log2[ (1+y)/y ]
Then (1 + y) / Y > 0
That is, y (y + 1) > 0
The solution is y < - 1 or Y > 0
That is, the value range of the original function is (- ∞, - 1) ∪ (0, + ∞)
This method is called inverse function method

Y = 1 / (x power of 2 - 1) range

Power X of 2 > 0
Power X of 2 - 1 > - 1
1 / (x power of 2 - 1) < - 1 or 1 / (x power of 2 - 1) > 0
Therefore, the value range is (- ∞, - 1) ∪ (0, + ∞)

Verify that 1-2sinxcosx / Cos2 power x-sin2 power x = 1-tanx / 1 + TaNx

(1-2sinxcosx)/(cos ² x-sin ² x)=(cos ² x+sin ² x-2sinxcosx)/(cos ² x-sin ² x)=(cosx-sinx) ²/ (cosx SiNx) (cosx + SiNx) = (cosx SiNx) / (cosx + SiNx) (numerator and denominator divided by cosx) = (cosx / cosx SiNx /

Given SiN4 power x-cos4 power x = - four fifths, then sin2x= For the specific problem-solving process, please don't skip steps,

SiN4 power x-cos4 power x = - four fifths
sin^4x-cos^4x=-4/5
(sin^2x+cos^2x)(sin^2x-cos^2x)=-4/5
1*(sin^2x-cos^2x)=-4/5
sin^2x-cos^2x=-4/5
cos^2x-sin^2x=4/5
Cos2x = 4 / 5. Cosine double angle formula
∴sin2x=±3/5