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1 + 3 / 4x = 2 solution equation
1+3/4x=2
3/4x=21
x=4/3
A function of the form y = KX (a constant of K ≠ 0) is called a positive proportional function, and its image is______ ?
A line passing through the origin
If the value of the independent variable of the positive proportion function increases by 7 and the value of the function decreases by 4, then when x = 4, the value of Y is obtained
Thank you
The analytic expression of positive proportion function is
y=kx (1)
(k is a constant and K ≠ 0), where x is an independent variable and Y is a function of X
By increasing the value of independent variable by 7, the value of function decreases by 4
y-4=k(x+7) (2)
From (1) (2), we can get k = - 4 / 7
So when x = 4, y = - 4 / 7 * 4 = - 16 / 7
The monotone increasing interval of function f (x) = log5 (2x + 1) is______ ．
In order to make the analysis of function meaningful, then 2x + 1 > 0, so the domain of definition of function is (- 12, + ∞). Because the inner function u = 2x + 1 is an increasing function and the outer function y = log5u is also an increasing function, the function f (x) = log5 (2x + 1) increases monotonically in the interval (- 12, + ∞), so the monotonic increasing interval of function f (x) = log5 (2x + 1) is (- 12, + ∞), so the answer is: (- 12, + ∞)
Finding monotone decreasing interval of function y = log5 (x ^ 2-2x-3)
Detailed process
Since y = log5 (x) is a monotone increasing function, the monotone decreasing interval of y = log5 (x ^ 2-2x-3) is the monotone decreasing interval of function g (x) = x ^ 2-2x-3
G (x) = (x-1) ^ 2-4, monotone decreasing interval is (infinite, 1)
The domain is x ^ 2-2x-3 > 0
(x-3)(x+1)>0
x>3,x
The monotone increasing interval of function f (x) = log5 (2x + 1) is (0, positive infinity)?
What is log5
I've got it. Who comes first
It should be that the monotone increasing interval of function f (x) = log5 (2x + 1) is (- 1 / 2, positive infinity)
Log in log5 (2x + 1) means logarithm,
So log 5 (2x + 1) is read as a logarithm with 5 as the base (2x + 1)
Does the function f (x) = log5 (2x + 1) have a monotone decreasing interval? How to change it?
No, if the base number is changed to one fifth, the decreasing interval will be (- 1 / 2, positive infinity); or 2x + 1 will be changed to twice the square of x plus 1, and the function will decrease monotonically at (negative infinity, 0)
If f (x) = x (ex + 1) + 12x2, then the monotone increasing interval of F (x) is______ ．
Let f ′ (x) = ex + 1 + xex + x = (ex + 1) (x + 1), Let f ′ (x) > 0, then we get x ＞ - 1, the monotone increasing interval of function f (x) is (- 1, + ∞), so the answer is: (- 1, + ∞)
If the function f (x) = a ^ & # 178; - 2x (where a > 0 and a ≠ 1) has the maximum value on R, then it satisfies the value range of loga (x-3) > 0
Are you sure the function is written correctly? It should be a ^ (2-2x). In this case, there is a solution
The solution is as follows: the maximum value of the function is only 0
What does the discerning radical of quadratic function mean
Ax ^ 2 + BX + C = 0ax ^ 2 + BX = - CX ^ 2 + BX / a = - C / ax ^ 2 + 2 * x * (B / 2a) + (B / 2a) ^ 2 = - C / A + (B / 2a) ^ 2 (x + B / 2a) ^ 2 = (b ^ 2-4ac) / 4A ^ 2x + B / 2A = ± √ (b ^ 2-4ac) / 2aX = [- B ± √ (b ^ 2-4ac)] / 2A make the discriminant △ = B ^ 2-4ac when △ 0, there are two roots
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