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The function y = (1 / 2) ^ x, the value range of x > 0 is?
The function y = (1 / 2) ^ x is a decreasing function,
When x = 0, y = 1
The range of x > 0 is 0
How to find the derivative of LN [f (x)]
Ln [f (x)] regards f (x) as an unknown quantity. From the formula, we get that the derivative is 1 / F (x)
The function y = log (SiNx - √ 3cosx) x belongs to the range of [π / 2, π]
【0,lg2】
Detailed explanation
sinx-√3cosx=2sin（x-π/3）
∵ x belongs to [π / 2, π], and∵ X - π / 3 belongs to [π / 6,2 π / 3]
In this interval, the value of SiNx - √ 3cosx is always greater than zero, and the maximum value is 1, and the minimum value is 1 / 2
The maximum value of y = LG (SiNx - √ 3cosx) is LG2 and the minimum value is 0
How to calculate the derivative of F (x) = - x-ln (- x)?
I know the answer is f '(x) = - 1 - (1 / x), but what is the specific process?
Because (- x) '= - 1
Also, let u (x) = - X
Then ln (- x) = ln (U (x))
(ln(-x))'=(lnu(x))' * u'(x)=(1/u) * (-1)=(1/-x) * (-1)=(1/x)
So f '(x) = - 1 - (1 / x)
Solve the equation 1.5x + 18 = 3.2X note that there is no multiplier sign
3.2x-1.5x=18
1.7x=18
x=18÷1.7
X = 10 and 10 out of 17
5.2x-0.1x4 = 4.8 find the equation 0.1x4, which is the multiplication sign
5.2x-0.1x4=4.8
5.2X=5.2
X=1
ALL
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