When f (x) = ex-ax-1, if f (x) increases monotonically in the domain R, then the value range of a is______ .

When f (x) = ex-ax-1, if f (x) increases monotonically in the domain R, then the value range of a is______ .

∵f（x）=ex-ax-1，∴f′（x）=ex-a，
∵ f (x) increases monotonically in the definition domain R, ᙽ ex-a ≥ 0 holds,
That is, a ≤ ex, ∵ ex ﹥ 0, ᙽ a ≤ 0
So the answer is: (- ∞, 0]

Given the function f (x) = ax-a / x-2lnx (a > 0), if the function f (x) is a monotone function in its definition domain, find the value range of A

Derivative f '(x) = a + A / x ^ 2-2 / x = (AX ^ 2-2x + a) / x ^ 2
If f (x) is a monotone function in its definition domain, it is necessary to make f '(x) ≥ 0 or F' (x) ≤ 0
1) If f '(x) ≥ 0 holds, then ax ^ 2-2x + a ≥ 0 holds;
For the curve y = ax ^ 2-2x + A, the opening is upward because a > 0;
When △ = 4-4a ^ 2 = 4 (1-A ^ 2) ≤ 0, y ≥ 0 is always true. When 1 ≤ a ^ 2, a ≥ 1 is obtained
When △ = 4-4a ^ 2 = 4 (1-A ^ 2) > 0, it is 0

The definition of the interval (x) of (x + 2) = (2) is given

ax^2+(a-2)x+1/4>0.
Start to discuss the situation of A
1、 If a > 0, then △ = (A-2) * (A-2) - a = a ^ 2-5A + 4

The monotone function f (x) satisfies f (AX + 3) = x, where a > 0. If the definition domain of F-1 (x) is [- 7 parts of a, 3 parts of a]. Find the analytic formula and definition domain D of F (x)

Because f (AX + 3) = X
So let: ax + 3 = t, so x = (T-3) / A
So f (x) = (x-3) / A
Let f (x) = (x-3) / a = y
So x = ay + 3
So F-1 (x) = ay + 3
Because the range of the inverse function is the domain of the original function
So a * (- 7 / a) + 3 = - 4
a*(3/a)+3=6
So the definition domain of the original function d = [- 4,6]

Given that the function f (x) = 3x, and f (a + 2) = 18, G (x) = 3ax-4x, the definition domain is [0,1] (1) Find the analytic formula of G (x); (2) find the range of G (x); (3) judge the monotonicity of G (x) and prove it by definition

1. Find a first
3(a+2)=18 a=4
Substituting a = 4 into G (x) gives g (x) = 8x
2. The definition domain is [0,1]
The range is [0,8]
3. Increasing function
Proof: set x1

Given that the definition domain of the function f (x) = x ^ 2-3x + 2 is (- ∞, 3 / 2), find the inverse function of function f (x)

y=f(x)=x^2-3x+2=(x-3/2)^2-1/4
X < 3 / 2, so Y > - 1 / 4
y+1/4=(x-3/2)^2
x<3/2,|x-3/2|=3/2-x
√(y+1/4)=√(x-3/2)^2=3/2-x
x=3/2-√(y+1/4)
So the inverse function is y = 3 / 2 - √ (x + 1 / 4), x > - 1 / 4

The definition domain of inverse function of y = (X-2) / X (x > 2) We need to explain the formula in detail The liberal arts student said that Alexander was a blockhouse in science

The definition domain of inverse function is the range of original function
y=(x-2)/x=1-2/x
x>2 0<2/x<1 1-1<1-2/x<1-0 0<1-2/x<1
The range of value of 0 primitive function is (0,1), and the range of definition of inverse function is (0,1)

The definition domain of inverse function of function y = x / X-2 (x 〉 2) is? We need answers urgently

Y = 1 + 2 / X-2, its range is 1, the definition domain of positive infinite inverse function is 1, positive infinity

It is proved that the monotonicity of the original function and the inverse function are the same It is known that y = f (x) is an increasing function on [a, b], It is proved that y = F-1 (x) is an increasing function on [f (a), f (b)] The beginning of the problem solving process has been given: If we take x1, X2 ∈ [f (a), f (b)], then there exists x'1, X'2 ∈ [a, b], such that f (x'1) = x1, f (x'2) = x2 Please help me finish this problem,

[proof]
If we take x1, X2 ∈ [f (a), f (b)] and X1 has x'1, X'2 ∈ [a, b], such that f (x'1) = x1, f (x'2) = x2
Because f (x) is an increasing function in [a, b]
So the larger the argument
According to the properties of X1 and inverse function, F-1 (x1) = X1 ', F-1 (x2) = x2'
Therefore, F-1 (x1) - F-1 (x2) = X1 '- x2' < 0
F-1 (x1) so function F-1 (x) is also an increasing function in [f (a), f (b)]

If the function y = f (x) (the definition field is D, the range is a) and the function y = f (x) has monotonicity on the definition field D, then does the function y = f (x) have an inverse function? (2) Are functions with inverse functions necessarily monotonic? (2) if it is not certain, can you give a counterexample (not monotonic function)?

(1) Because monotone function is one-to-one, it must have inverse function
(2) For example, f (x) = 1 / x, although it is monotone decreasing on (- ∞, 0) and (0, ∞), it does not have monotonicity on (- ∞, 0) U (0, + ∞),
Its inverse function is f ^ (- 1) (x) = 1 / X