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Ln function domain The domain of the function z = ln [(25-x2-y2) (x2 + y2-4)] is? The two after X and y are squares
four
Is there any difference between an open interval and a closed interval when the domain of definition or value of a function is positive or negative infinity?
Is there any difference between an open interval and a closed interval when the domain of definition or value of a function is from negative infinity to positive infinity?
When the function interval is infinite, there is no closed interval
What is the most obvious difference between a range and a domain?
B sweat··
Function y = f (x)
Definition field refers to the value range of X; value field refers to the value range of Y, that is, X takes all the values of the definition field, and the corresponding Y values form a set of value fields
How to express the mathematical definition field and value field?
{x|.}
{y|.}
Y = x + radical (1-x ^ 2) range
1-x ^ 2 > = 0 and
Difference between definition field and value field
For example, y = x ^ 2 + X + 1
The definition field is the value range of X. if R is not specified, some are based on whether the formula is meaningful and absolute. For example, y = root sign (x + 3) because the formula under the root sign is greater than or equal to 0, the definition field x is greater than or equal to - 3, and the value field y is greater than or equal to 0
Range refers to the value range of Y, y = x ^ 2 + X + 1 = (x + 1 / 2) ^ 2 + 3 / 4 ≥ 3 / 4 range, y is greater than or equal to 3 / 4 domain, X is r
Derivative of y = (2x-1) ^ 10
process
This is a composite function of a power function, and its derivative is:
y'=10*(2x-1)^9*[(2x-1)]'
=10(2x-1)^9*2
=20(2x-1)^9.
Find y = (2x + 1) ^ 10 derivative y '
y‘=10×2×(2x+1)^9=20(2x+1)^9
Derivative of y = (x-1) (x + 1) (x + 3)
Derivative of y = (x-1) (x + 1) (x + 3)
y'=(x+1)(x+3)+(x-1)(x+3)+(x-1)(x+1)
=X-square + 4x + 3 + x-square + 2x-3 + x-square-1
=3x square + 6x-1
Given LNX = 2 + ln (2 / x), find the answer of X, x = 2 × e under the root sign, find the analysis
Inx=2+In2-Inx
2Inx=2+In2
Inx=(2+In2)÷2=1+1／2In2=1+In√2=Ine+In√2=In√2e
∴x=√2e,Ine=logeE=1
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