Let a be greater than 0, f (x) = e ∧ 2 / A + A / E ∧ X be the even function defined on R To prove that f (x) is an increasing function on (0, infinity)

Let a be greater than 0, f (x) = e ∧ 2 / A + A / E ∧ X be the even function defined on R To prove that f (x) is an increasing function on (0, infinity)

If f '(x) = 1-1 / e ^ x, then Lim X - ∞ 1 / e ^ x = 0, we can prove that 1 / e ^ x must be less than or equal to 0, so 1-1 / e ^ x must be greater than 0, because the derivative is greater than 0, so it is an increasing function

It is known that the function f (x) = a + (1 / 1 + 2x) defined on R is an odd function
It is known that the function f (x) = a + (1 / 1 + 2x) defined on R is an odd function (1) find the value of a (2) prove that the function f (x) is a decreasing function on R (3) if for any t ∈ R: inequality f (t2-2t) + F (2t2-k)

Solution;
(1) If f (x) defined on R is an odd function, then f (0) = a + 1 / (1 + 2 ^ 0) = 0, a = - 1 / 2 is obtained
(II)f(x)=-1/2+1/(1+2^x)
Let X1 > x2
f(x1)-f(x2)=1/(1+2^x1)-1/(1+2^x2)=(2^x2-2^x1)/[(1+2^x1)(1+2^x2)]
Because X1 > X2, there are 2 ^ X1 > 2 ^ x2,1 + 2 ^ X1 > 0,1 + 2 ^ x2 > 0
So f (x1) - f (x2)

Let a ∈ R, f (x) = ax-2x-2a. If the solution set of F (x) > 0 is a, B = {x 1 ＜ x ＜ 3}, a ∩ B =, find the value range of real number a?

∩ B = empty set, B ≠ empty set ∪ a = empty set or a = (- ∞, 1] ∪ [3, + ∞) ∪ △ 4 + 8A ^ 2 > 0 ∩ according to the image, the solution set of function f (x) cannot be empty set ∪ a = (- ∞, 1] ∪ [3, + ∞) ∪ a > 0. I'd like to help you!

Given f (x) = ax ^ 2 - x + 2a-1, let H (x) = f (x) / x, if the function H (x) is an increasing function in the interval 1 ＜ = x ＜ = 2, find the value range of real number a
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F (x) = ax & # 178; - x + 2a-1=
ax²-x+2a-1,x≥0;
ax²+x+2a-1,x0;
When ax + 1 + (2a-1) / x, x0, H (x) = x + (2-1 / a) / X-1
So let H (x) increase on [1,2]
Only √ (2-1 / a) ≤ 1
The solution is a ≤ 1

Let a belong to R, and the domain of the function y = LG (AX & # 178; - 2x-2a) be a, B = {x 1 ＜ x ＜ 3}. If a ∩ B ≠ empty set, the value range of real number a is obtained

Suppose a ∩ B is an empty set
Then ax ^ 2-2x-2a

Given the function f (x) = x ^ 3 + ax ^ 2 + 3 / 2x + 3 / 2a, and f '(- 1) = 0.1, find the value 2 of a, and find the maximum value of F (x) on [- 1,0]
Let f (x) = x ^ 3 + ax ^ 2 + (3 / 2) x + (3 / 2) a, and f '(- 1) = 0
1, find the value of A
2, find the maximum value of function f (x) on [- 1,0]

(1)
f(x)=x^3+ax^2+（3/2）x+（3/2）a
f'(x)=3x^2+2ax+3/2
f'(-1)=3*(-1)^2+2a*(-1)+3/2=0
3-2a+3/2=0
2a=3+3/2=9/2
a=9/4
f'(x)=3x^2+2*9/4x+3/2
=3x^2+9x/2+3/2
=3(x^2+3x/2+9/16)-27/16+3/2
=3(x+3/4)^2-3/16
When x = - 3 / 4, there is a maximum value, which is not in the interval [- 1,0]
So the minimum value of function f (x) on [- 1,0]
Is f (- 1) = (- 1) ^ 3 + 9 / 4 * (- 1) ^ 2 + 3 / 2 * (- 1) + 3 / 2 * 9 / 4
=-1+9/4-3/2+27/8
=25/8
The maximum value of function f (x) on [- 1,0]
It is f (0) = (3 / 2) * 9 / 4
=27/8

If the function f (x) = AX2 + 2x + A + 3 satisfies f (1 + x) = f (1-x), then the value of a is______ ．

∵ function f (x) = AX2 + 2x + A + 3, satisfying f (1 + x) = f (1-x), ∵ the symmetry axis of function image is x = - 1A = 1, a = - 1 is obtained, so the answer is: - 1

Given that the function f (x) = ax / 2x-1 satisfies f [f (x)] = x, find the value of real number a

F (x) = ax / (2x-1), f (f (x)) = a [ax / (2x-1)] / [2aX / (2x-1) - 1] = A & # 178; X / [2aX - (2x-1)] = x; simplify a & # 178; X = x [2aX - (2x-1)] → 2 (A-1) x & # 178; + (1-A & # 178;) x = 0; for any x, then A-1 = 0 and 1-A & # 178; = 0, so a = 1

It is known that the maximum value of the function y = - x ^ 2 + ax-a / 4 + 1 / 2 in the interval [0.1] is g (a)
Write the expression of G (a)
Finding the minimum of G (a)

y=-x^2+ax-a/4+1/2
=-(x-a/2)^2+a^2/4-a/4+1/2
The opening of the function is downward and has the maximum symmetry axis X = A / 2
(1)

If the two zeros of quadratic function y = x2 + BX + C are - 1 and 2 respectively, then the solution set of inequality f (x) < 0 is______ ．

The two zeros of quadratic function y = x2 + BX + C are - 1 and 2 respectively, and the opening of quadratic function is upward. The solution set of F (x) < 0 is (- 1,2), so the answer is: (- 1,2)