The square of x = x + 3

The square of x = x + 3

Direct observation shows that any term of the function y is nonnegative and has a minimum value at x = 0, and the value range is [3, + ∞]

Find the value range of x power of function y = (1 / 4) - (1 / 2) x power + 1 (x in - 3 and 2) Can give me an answer as soon as possible, It's a closed area. Your maximum is wrong... The correct answer is 57.

Ha ha, wrong reading:
X power of y = (1 / 4) - (1 / 2) + 1
y=(1/2)^2x-(1/2)^x+1
=[(1/2)^x-1/2]^2+3/4
-3

Find the value range of the x power of the function y = radical 16-4

Because y = √ (), y is y ≥ 0 first, and since the root must be positive, the x power of 4 is always greater than zero, so the maximum value of x power of 16-4 is 16, but it cannot be obtained
So the range of Y is [0,4]

Find the value range of x power of function y = 2 + x power of 2 / 1

Tip: the value range of a function is the domain of its inverse function
y=2^x/(2^x +1)
Organize, get
2^x=y/(1-y)
2^x>0
y/(1-y)>0
y/(y-1)

The function y = 10 to the power of X, where x belongs to the range of n *-

｛y|y=10^x,x∈N*｝

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

y=1/(2^x-1)
Moving to y × 2 ^ X - y = 1
2^x=(1+y)/y
Logarithm of 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 to say, the value range of the original function is (- ∞, - 1) ∪ (0, + ∞)
This method is called inverse function method

The range of y = 1 / (x-1 of 2)

The power of 2 is greater than 0
Power X of 2 - 1 > - 1
1 / (x-1 of 2) < - 1 or 1 / (x-1 of 2) > 0
So the range is (- ∞, - 1) ∪ (0, + ∞)

To get the image of the function y = sin (4th power) x-cos (4th power) x, just change the function y = 2sincos?

Let's use a ^ B to denote the B power of A. = = = = = = = = = because (SiN x) ^ 4 - (COS x) ^ 4 = [(SiN x) ^ 2 + (COS x) ^ 2] [SiN x ^ 2 - (COS x) ^ 2] = - cos 2x = sin (2x - π / 2) = sin 2 (x - π / 4), and because 2 SiN x cos x = sin 2x, y is obtained

The third power of X + XY sin (π y) = O. find the derivative of Y I hope the sooner the better. I'm in urgent need

The derivation of Y on both sides of the equation is obtained
X-πcos(πY)=O
therefore
cos(πY)=X/π
Y={arccos(X/π)}/π
The derivative of Y
Under y '= - 1 / π * {1 / radical [1 - (x / π) ^ 2]}

What is the derivative of y = SIN3 to the power of X?

Let SiNx = a, that is, y = a ^ 3, and derive y = 3sin ^ 2xcosx