The maximum and minimum values of F (x) = (a + SiNx) (a + cosx), (a > 0)

The maximum and minimum values of F (x) = (a + SiNx) (a + cosx), (a > 0)

f(x)
=a^2+sinxcosx+a(sinx+cosx)
from
1+2sinxcosx=(sinx+cosx)^2
Let SiNx + cosx = t, t ∈ [- sqrt2, sqrt2]
g(t)=a^2+(t^2-1)/2+at
=1/2t^2+at+a^2-1/2
The axis of symmetry is x = a
0

F (x) = 1-sinx-cosx + 2sinx cosx Max and min

Let SiNx + cosx = t (- √ 2)

Given the function f (x) = 3 + log2x, X belongs to [1,4], G (x) = f (x ^ 2) - [f (x)] ^ 2
Explain why x belongs to [1,2]

The solution is given by the function f (x) = 3 + log2x, X belongs to [1,4]
The scope of action of the law of knowing correspondence f is [1,4]
So in the function g (x) = f (x ^ 2) - [f (x)] ^ 2
Where 1 ≤ x ^ 2 ≤ 4 and 1 ≤ x ≤ 4
That is - 2 ≤ x ≤ - 1 or 1 ≤ x ≤ 2 and 1 ≤ x ≤ 4
The solution is 1 ≤ x ≤ 2
So the definition field of function g (x) is [1,2]

log2x=4 ,x^-1/2=

log2x=4 ,x=16
x^-1/2=1/4

Given that f (x) = 2 + log2x, 1 = < x ＜ = 16, then the maximum value of function y = [f (x)] squared + F (x squared) is?
Detailed steps
The standard answer is 22

y=[f(x)]^2+f(x^2)
=4+4log2x+(log2x)^2+2+log2x^2
=(log2x)^2+6log2x+6
Let t = log2x, 4 ≥ t ≥ 0 from 16 ≥ x ≥ 1
y=(t+3)^2-3
When t = 4, i.e. x = 16
Y has a maximum of 46

Given the function f (x) = log2x. F (x, y) = x + Y2, then f [f (1 / 4x1) equals?

Should it be f (f (1 / 4), 1?
f(1/4)=log2(1/4)=-2
F(f(1/4),1)=F(-2,1)=-2+1^2=-2+1=-1

Let the inverse function of the function y = f (x) be y = f ^ - 1 (x) and the image of y = f (1-3x) passes through the point (1 / 3,1), then the image of the function y = f ^ - 1 (x) passes through the point (1 / 3,1)

Image passing point (1,0)

Given that function y = (1 / 2) x + m and function y = nx-1 / 3 are reciprocal functions, the value of M and N can be obtained

Because they are inverse functions, the first functions X and y are exchanged to get x = (1 / 2) y + m, that is, y = 2x-2m, so 2 = n, 2m = 1 / 3, M = 1 / 6

=If Mn = 16 (m, n ∈ positive real number), then F-1 (m) + F-1 (n)=
also
Y = Log1 / 2 x x belongs to the (0,8] range

f(x)=2^(x+3)
f^(-1)(x)=log2(x)-3
f-1(m)+f-1(n)
=log2(m)-3+log2(n)-3
=log2(mn)-6
=log2(16)-6
=-2

Given the function f (x) = log2 (1 + X / 1-x), prove f (x1) + F (x2) = f [(x1 + x2) / (1 + x1x2)

f(x1)+f(x2)=log2[(1+x1)/(1-x1)]+log2[(1+x2)/(1-x2)]
=log2[(1+x1)(1+x2)/(1-x1)(1-x2)]
f[(x1+x2)/(1+x1x2)]
=log2[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)]
It is to prove that (1 + x1) (1 + x2) / (1-x1) (1-x2) = [1 + (x1 + x2) / (1 + x1x2)] / [1 - (x1 + x2) / (1 + x1x2)
1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)
=(x1x2+x1+x2)/(x1x2-x1-x2)
And (1 + x1) (1 + x2) / (1-x1) (1-x2)
=(1+x1x2+x1+x2)/(1+x1x2-x1-x2)
Obviously this equation doesn't hold, so the title is wrong