Find y = log2 (x ^ 2-x) range

Find y = log2 (x ^ 2-x) range

x²-x
=(x-1/2)²-1/4≥-1/4
That is, true numbers can get all positive numbers
So the range is r

What is the range of the function y = log2 (x) + 2?

Log2 (x) has only R
So the range of Y is also r
That is (- ∞, + ∞)

If f (x) = x & # 178; - 2ax-3 has an inverse function on [1,4], find the value range of A

If f (x) = x & # 178; - 2ax-3 has an inverse function on [1,4], then f (x) = x & # 178; - 2ax-3 must be a monotone function on [1,4]
Then a = 4

If the definition field of function f (x) = 1 / (e ^ x + ax) is the value range of R for a

The definition field of function f (x) = 1 / (e ^ x + ax) is r, that is, e ^ x + AX = 0 has no solution. ∵ x = 0 has no solution, that is, the image of y = e ^ X and the image of y = - ax have no common point. First, find the tangent line of y = e ^ x passing through the origin. Let the tangent point be (x0, e ^ x0) y '= e ^ x ∵ k = e ^ x0 ∵ the tangent equation be y-e ^ x0 = e ^ x0 (x-x0) ∵ the line passes through the origin ∵ 0-E ^ x0

Let f (x) = (e ^ x) / (x ^ 2 + ax + a) be the value range of R for a

The definition field of function f (x) = (e ^ x) / (x ^ 2 + ax + a) is r
X ^ 2 + ax + A is not equal to zero
That is, the image of G (x) = x ^ 2 + ax + A has no intersection with the X axis
Since a > 0, as long as Δ

Given that the domain of function f (x) is x ∈ [- 12,32], find the domain of G (x) = f (AX) + F (XA) (a > 0)

Let μ 1 = ax, μ 2 = Xa, where a ＞ 0, then G (x) = f (μ 1) + F (μ 2) and μ 1, μ 2 ∈ [- 12, 32]; - 12 ≤ ax ≤ 32-12 ≤ Xa ≤ 32 {- 12a ≤ x ≤ 32a-a2 ≤ x ≤ 32A. ① when a ≥ 1, the solution of the inequality system is - 12a ≤ x ≤ 32A; ② when 0 < a < 1, the solution of the inequality system is - A2 ≤

If f (x) = x-1-lnx, then the minimum value of F (x) is_____ .

The minimum value of F (x) is 0

Given the function f (x) = x * LNX (1), find the minimum value of function f (x)
(2) If for all x ∈ (0, + ∞),
If f (x) ≤ X & # 178; - ax + 2 is constant, find the value range of real number a;
(3) Try to judge whether the function y = lnx-1 / ex + 2 / ex has zeros? If so, find out the number of zeros

(1) If f (x) ' = LNX + 1 and the reciprocal of F (x) ' is 0, then LNX = - 1, x = 1 / E, so the minimum value is f (1 / E) ' = - (1 / E)
(2) Xlnx & lt; = x ^ 2-ax + 2 holds in X & gt; 0
Solve a & lt; = (x ^ 2 + 2-xlnx) / x = x + 2 / x-lnx, and the right side of the equation is g (x)
Let's find the minimum value of G (x) g & # 39; (x) = 1-2 / x ^ 2-1 / x = (x ^ 2-x-2) / x ^ 2 = (X-2) (x + 1) / x ^ 2, then the minimum point x = 2g (2) = 2 + 1-ln2 = 3-ln2 is the minimum value, so there is a & lt; = 3-ln2
（3）
There is no zero

Higher numbers, y = e ^ X and x = e ^ y are both inverse functions of y = LNX and x = LNY
Y = e ^ X and x = e ^ y are both inverse functions of y = LNX and x = LNY. Why is it right

You know X and y are just an algebra, which means a variable, so it's better to understand them

Finding the inverse function of function y = (1 / 2) LNX

lnx=2y
x=e^2y
So when the inverse function is y = e ^ (2x), X ∈ R