Is f (x) = 1 and G (x) = x to the power of 0 the same function

Is f (x) = 1 and G (x) = x to the power of 0 the same function

No, the former domain is all real numbers, while the latter cannot take 0

(1) it is proved that the function f (x) = x power of E + x power of E is an increasing function on [0, + ∞]
(2) prove that the function y = xsinx + cosx is an increasing function in the interval (3 π / 2,5 π / 2)

1）f'(x)=e^x-e^(-x)=[e^(2x)-1]/e^x
∵ x ∈ [0, + ∞) ∵ e ^ (2x) - 1 ≥ 0 ∵ f '(x) ≥ 0, so it is increasing
2）y'=sinx+xcosx-sinx=xcosx
∵x∈(3π/2,5π/2),
∴cosx≥0
∴y'≥0
It is an increasing function

It is known that y = f (x) is an odd function defined on R. when x ≤ 0, f (x) = 2x + X2 (1) find the analytic expression of F (x) when x > 0; (2) if the equation f (x) = 2A2 + a about X has three different solutions, find the value range of A

(1) Let X be greater than 0, then - x ＜ 0, then - f (- x) = - 2x + (- x) 2 = x2-2x. ∵ f (x) is an odd function, ∵ f (x) = - f (- x) = 2x-x2. Therefore, when x ＞ 0, f (x) = 2x-x2. (2) from (1), y = f (x) has a maximum of 1, and a minimum of - 1 ∵ the equation f (x) = 2A2 + A has three different solutions, ∵ - 1 ＜ 2A2 + a ＜ 1. ∵ 1 ＜ a ＜ 12

Y = [root (2x-x Square) / LG (2x-1)] + {cubic [(x-1) / 3 | x | - 2]} to find the domain of definition

5 / 4 × (X-2 / 5) = 2
X-2 / 5 = 2 △ 5 / 4
X-2 / 5 = 8 / 5
X = 2 / 5 + 8 / 5
x=2

Given that the domain of F (x squared) is [- 1,1], then the domain of F (x) is (), and the domain of F (2 to the x power) is ()
Given that the domain of F (x squared) is [- 1,1], then the domain of F (x) is (), and the domain of F (2 to the x power) is ()
Then the domain of F (x) is ([0,1] is a closed interval, and if the domain of F (2) is (- ∞, 0), then 0 cannot be obtained.

The domain of F (x square) is [- 1,1], then the domain of F (x) is ([0,1]), and the domain of F (2 to the x power) is ((- ∞, 0])

Given that the function FX = 2x is an increasing function from negative infinity to positive infinity, we can find the direction of real number a
It is known that the square of the cubic power of function FX = 2x-x + ax + 1-A is an increasing function from negative infinity to positive infinity. 1) find the value range of A. 2) if the zero point of function belongs to interval (0,1), find the value range of real number a

Incrementing over R
Then f '(x) = 6x & # 178; - 2x & # 178; + a > 0
So the discriminant △

It is known that FX = a (x power of A-X power of a) / (square of A-2) is an increasing function on R, and the range of a is obtained

A:
F (x) = a * [a ^ x-a ^ (- x)] / (a ^ 2 - 2) is a monotone increasing function on R
Derivation:
F '(x) = a * (LNA) * [a ^ x + A ^ (- x)] / (a ^ 2-2) > = 0
Because: a > 0
So: A ^ x > 0, a ^ (- x) > 0 holds
So: (LNA) / (a ^ 2-2) > = 0 holds
1) 00, a > √ 2 or a √ 2
To sum up, 0

Given; X + 1 / x = 3, find 1. The second power of X + 1 / 2 of X; 2. The second power of [X-1 / x]

If x + x 1 / 2 = 3 and both sides square at the same time, we can get the quadratic power of X + x 1 / 2 + 2 = 9. Therefore, the quadratic power of X + x 1 / 2 = 7. The quadratic power of [x-x 1 / 2] = [x + x 1 / 2] - 4 = 5

If f (x) = (x power of 2 + 1), then f (f (0))=

f(0)=1/(2^0+1)=1/2
f(f(0))=f(1/2)=1/(√2+1)=√2-1

Decomposition factor: the square of (a + b) - the square of 4C

( a + b )" - 4c"
= ( a + b )" - ( 2c )"
= ( a + b - 2c )( a + b + 2c )