To their surprise Conflict with Pay more attention I feel stressed Change your mind How do you say these phrases in English? Some children like to sleep with the light on Some children prefer to sleep()?

To their surprise Conflict with Pay more attention I feel stressed Change your mind How do you say these phrases in English? Some children like to sleep with the light on Some children prefer to sleep()?

what surprises them is
have conflict with
pay more attention
feel the pressure
change idea
Sleep with light on
To their surprise
conflict with
pay more attention to
feel pressure
change one's idea
with light on
to their surprized
have conflicts
pay more attention to
feel pressured
change one's mind
with lights on
"Both How do you say this phrase in English
Both are...
english phrases
1 pen pal 2 from. 3 next to 4 on the left / right 5 in front of 6 turn left / right 7 along. Go straight 8 to 9 house with garden 10 a little 11 keep quiet 12 in the daytime 13 at night 14 give someone something
1 penfriend2 come from3 be closed to4 on the left/right5 in front of6 turn left/right7 go along with8 arrive at9 the house with garden10 a bit11 keep quiet12 in the morning13 at night14 give sb sth
Both are translated into English phrases
both of.
The simplest is both
both of …… Are
Then another change can be made
both and
A proof of periodic function
The answer is yes, but I don't know how one of the steps is derived
Let f (x) be defined in (- ∞, + ∞) and satisfy f (x + π) = f (x) + SiNx. It is proved that f (x) is a periodic function with period of 2 π.
The proof idea is f (x + 2 π) = f (x)
The answer is f (x + 2 π) = f [(x + π) + π] = f (x + π) + Sina (x + π) = [f (x) + SiNx] + (- SiNx) = f (x), where f [(x + π) + π] = f (x + π) + sin (x + π). I don't know how to deduce this part of sin (x + π)
Ok... I found it myself. I can take x = x + π in it. Orz... It took me a long time to find out
It's very simple, let x + π = t
Then f [(x + π) + π] = f (T + π) = f (T) + Sint = f (x + π) + Sina (x + π)
Let y = f (x), X belong to R, and its image is symmetric with respect to x = a, x = B, (a is less than B).
It is proved that y = f (x) is a periodic function and its period is obtained.
The explanation is as follows
If x = A and x = B are symmetric, then f (x) = f (A-X) f (x) = f (b-X) holds, because x belongs to R, if - x is brought into f (x), then,
Let f (- x) = f (a + x) = f (B + x) let x choose x = x-a, then f (a + x-a) = f (B + x-a), that is, f (x) = f (x + B... expansion
Let y = f (x), X belong to R, and its image is symmetric with respect to x = a, x = B, (a is less than B).
It is proved that y = f (x) is a periodic function and its period is obtained.
The explanation is as follows
If x = A and x = B are symmetric, then f (x) = f (A-X) f (x) = f (b-X) holds, because x belongs to R, if - x is brought into f (x), then,
Let f (x + a) = x + B (x + a) = F-A (x + b) = F-A (x + a) = F-A (x + b) = F-A (x + b) = F-A (x + a) = F-A (x + b) = F-A (x + B)
The period defined by periodic function is b-a
This kind of problem must be done by definition. Put it away
If f (x + π) = f (x) + SiNx is satisfied, then, simply speaking, this function can be written as f (y + π) = f (y) + siny. When y = x + π, this function becomes f (x + 2 π) = f (x + π) + sin (x + π); that is, f [(x + π) + π] = f (x + π) + sin (x + π). The main purpose is to understand the connotation of the function. This x, y is only a variable, because it is defined in (- ∞, + ∞), x, y is just a name For example, when I was in primary school, I learned to use letters to express numbers. These are coherent knowledge
If f (x + π) = f (x) + SiNx is satisfied, then, simply speaking, this function can be written as f (y + π) = f (y) + siny. When y = x + π, this function becomes f (x + 2 π) = f (x + π) + sin (x + π); that is, f [(x + π) + π] = f (x + π) + sin (x + π). The main purpose is to understand the connotation of the function. This x, y is only a variable, because it is defined in (- ∞, + ∞), x, y is just a name For example, when we were in primary school, we used letters to express numbers. This is a coherent knowledge, but sometimes we ignore it
How to find the range of logarithmic function
First of all, the definition field is required, and then the value field is obtained according to the logarithm function image
The domain of definition of the inverse function of the function y = (x square of 2-1) / x square of 2 is
The domain of inverse function is the domain of original function
y=(2^x-1)/2^x=1-(1/2)^x
(1 / 2) ^ x belongs to (0, + infinity)
So the range is (- infinity, 1)
That is, the domain of inverse function is (- infinity, 1)
The domain of the inverse function is the domain of the original function, as long as the domain of the original function is determined.
y=(2^x－1)/(2^x)
=1－(1/2)^x
Considering that the range of (1 / 2) ^ x is 0 to infinity, then: y ∈ (- ∞, 1)
So, the domain of the inverse function of this function is (- ∞, 1)
Y = (x square of 2-1) / x square of 2
=1-1/2^x
1/2^x=1-y
2^x=1/(1-y)
x=log2 1/(1-y)=-log2 (1-y)
The inverse function is y = - log2 (1-x)
The domain is 1-x > 0
The solution is X
How to judge whether a function is a periodic function
Help to prove the following question: 2cos (x / 2) - 3sin (x / 3) is a periodic function, its period is? Online, etc., for God!
If the function is periodic and the period is t (T > 0), then f (x) = f (x + T) sin | x + T | = sin | x | when - T
Range of logarithmic function
y=log1/2 (8x-x^2 -7)
Find the domain, value, and monotone interval
If 8x-x ^ 2-7 > 0, then (1,4) should be a single increasing interval, and (4,7) should be a decreasing interval. (because a = 1 / 2 is a decreasing function in (0, positive infinity), and 8x-x ^ 2-7 is an increasing function after (1,4] is a decreasing function, and the decreasing function and the decreasing function are combined, so (1,4] should be the increasing interval of the whole function
8x-x^2-7>0, (x-1)(x-7)
If f (x) + F (- x) = 4 holds for everything, then F-1 (x) is the inverse function of y = f (x)
If f (x) + F (- x) = 4 holds for everything, then the value of f ^ - 1 (x-3) + f ^ - 1 (7-x)
Let x = 5 f ^ - 1 (x-3) f ^ - 1 (7-x) = f ^ - 1 (2) f ^ - 1 (2) and f (0) f (0) = 4 f (0) = 2, so f ^ - 1 (2) = 0, so the original formula = 0, 0 = 0