Find the function y = SiNx cosx + sinxcosx (0

Find the function y = SiNx cosx + sinxcosx (0

Let SiNx cosx = t, then
t＝√2sin(x-π/4),
∵x∈(0,π),∴x-π/4∈(-π/4,3π/4),
∴-√2/2＜sin(x-π/4)≤1,
∴-1＜t≤√2.
And the square of both sides, 1-2sinxcosx = T ^ 2
∴sinxcosx=(1-t^2)/2
∴y=sinx-cosx+sinxcosx=t+(1-t^2)/2
＝(-1/2)(t^2-2t)+(1/2)
=(-1/2)(t-1)^2+1
When t = 1, y (max) = 1;
When t = - 1, y (min) = - 1;
(t = - 1 cannot be taken, so y (min) = - 1 cannot be realized)
In conclusion, y ∈ (- 1,1],
That is, the maximum value is 1. The minimum value - 1 cannot be obtained (there is no minimum value)

Y = sinxcosx + SiNx + cosx max min
.
What is derivative? Let SiNx + cosx = t, then the square of sinxcosx = t plus 1 divided by 2, and the maximum value is obtained by quadratic function

Its derivative is
Cos2x + cosx SiNx, x = 1 / 2pi is zero, x = 1 / 4Pi
Answer: 1 / 2 + radical 2

Finding the maximum value of y = 2 (SiNx + cosx) - sinxcosx

Let a = SiNx + cosx = √ 2Sin (x + π / 4)
So - √ 2

The maximum value of the function y = sinxcosx + SiNx + cosx is___ ．

Let t = SiNx + cosx = 2Sin (x + π 4), then - 2 ≤ t ≤ 2 ﹣ sinxcosx = t2-12 ﹣ y = 12t2 + T-12 = 12 (T + 1) 2-1 (- 2 ≤ t ≤ 2) axis of symmetry t = - 1 ﹣ when t = 2, y has the maximum value of 12 + 2, so the answer is 12 + 2

It is known that sin ^ 2x + sin 2xcos x-cos 2x = 1,0

Sin ^ 2x what does that mean

The domain of y = lgx
I forgot all I learned

If x ∈ R, the domain of definition is R. also note that if y = LG (4-x0), the domain of definition is required, because 4-x ＞ 0, that is, X ＜ 4, so the domain of definition of function y = LG (4-x) is {X / X ＜ 4}

The domain of y = logx + 1 (5-x)

Logx, the domain is x > 0
1 / (5-x), the domain is x ≠ 5
So the domain of y = logx + 1 / (5-x) is: x > 0 and X ≠ 5

[[log3 (lgx)] = 1] [logx (8) = 3 / 2] [5 ^ lgx = 25,], how much is x equal to? 2

log3(lgx)=1
lgx=3^1=3
log10(x)=3
x=10³
x=1000
logx(8)=3/2
That is, x ^ (3 / 2) = 8
x=8^(2/3)=(2³)^(2/3)=2^2
x=4
5^lgx=25=5²
lgx=2
So log10 (x) = 2
x=10²=100

Is lgx = logx?

Lgx is a function with base 10 and logarithm X. LNX is a function with base E and logarithm X
There is no such writing method as logx, so we need to give a base

The range of y = log 3 x + log x 3 - 1

Y = log 3x + logx3 - 1 = log 3x + 1 / log3x - 1, if x ≥ 1, y ≥ 1, if 0