Given the function f (x) = 2acos & sup2; X + bsinxcosx, and f (0) = 2, f (π / 3) = 1 / 2 + √ 3 / 2

Given the function f (x) = 2acos & sup2; X + bsinxcosx, and f (0) = 2, f (π / 3) = 1 / 2 + √ 3 / 2

f(0)=2,f(π/3)=1/2+√3/2.
Easy to get a = 1, B = 2
f（x）=2cos²x+2sinxcosx
=sin2x+cos2x+1
=√2sin(2x+π/4)+1
The minimum positive period of function T = 2 π / 2 = π

1. Prove that f (x) = (a ^ x-1) x / (a ^ x + 1) [a > 0, a not = 1] is even function
2. It is known that a * B ^ x, f (4) = 648, f (5) = 1944
Estimated f (4.5)
Calculate f (4.5) and use the results to judge your estimate
3. It is known that f (x) = 3 ^ x, u and V belong to R
Write an equation about f (U) △ f (V) similar to f (U). F (V) = f (U + V). And prove your conclusion

1.f(-x)=[a^(-x)-1](-x)/[a^(-x)+1]
Multiply the numerator and denominator by a ^ x to get
f(-x)=(1-a^x)(-x)/(1+a^x)=(a^x-1)x/(a^x+1)=f(x)
So f (x) is an even function
2.a*b^4=648 a*b^5=1944
Estimated f (4.5) = (1944 + 648) / 2 = 1296
b=1944/648=3 b^0.5=1.732
F (4.5) = a * B ^ 4 * B ^ 0.5 = 648 * 1.732 = 1122, the estimation is not very good, but relatively close
3.f(u)÷f(v)=f(u-v)
Prove: F (U) = 3 ^ u, f (V) = 3 ^ v
f(u-v)=3^(u-v)
f(u)÷f(v)=3^u/3^v=3^(u-v)=f(u-v)

The range of function 1 / 3 ^ x2-2x-1 is? Monotone increasing interval? Monotone decreasing interval? Monotone decreasing interval?
Function (1 / 3) ^ (x ^ 2-2x-1)

x^2-2X-1=(x-1)^2-2 ≥-2
And it decreases in (- ∞, 1) and increases in (1, + ∞)
So the range of function (1 / 3) ^ (x ^ 2-2x-1) is y ∈ (0,9]
Monotone increasing interval is (- ∞, 1]
The monotone decreasing interval is (1, + ∞)

The problem of index function in senior one
Comparison: the size of (2x ^ 2 + 1) power of function a and (x ^ 2 + 2) power of function a
Process, speed
The power of (2x ^ 2 + 1) of function a and the size of (x ^ 2 + 2) and a > 1 of function a

t1=2x^2+1
t2=x^2+2
t1-t2=x^2-1
a>0,and a≠1
a>1
-1

A problem of exponential function
The range of x power of function y = 4 + x power of function 2-3

X power of y = 4 + x power of 2-3
=(2^x+1/2)^2-13/4
We know that R is monotonically increasing
So the range of Y is (- 3, + infinity)

The solution to the problem of finding the maximum value of the mathematical exponential function in grade one of senior high school
Let x ∈ [0,2], find the maximum value of function y = 4x-1 / 2 (x-1 / 2 is superscript) - 3 × 2x (x is superscript) + 5

Replace 4 with the square of 2, exponentially multiply it to the X-1 power of 2, and then change the element so that the x power of 2 is equal to t, and t belongs to 1 to 4. Finally, the maximum value is 5 / 2, and the minimum value is - 5

Given 2x2 + X ≤ (14) x − 2, the range of function y = x2-2x is obtained

∵ 2x2 + X ≤ (14) x − 2, ∵ 2x2 + X ≤ 24 − 2x, then x2 + X ≤ 4-2x, that is, X2 + 3x-4 ≤ 0, and the solution is - 4 ≤ x ≤ 1. ∵ y = x2-2x = (x-1) 2-1, we can see that the function decreases monotonically on [- 4,1], and the value range is [- 1,24]

A problem about the exponential function of mathematics in grade one of senior high school
The monotone increasing interval and monotone decreasing interval of 1-2 ^ (x + 1) + 2 ^ 2x under the function y = root sign are pointed out
Detailed process, please

1-2^(x+1)+2^2x=1-2*2^x+(2^x)^2=(2^x-1)^2
y=1-2^x(x0)
So y decreases monotonically on (negative infinity. 0) and increases monotonically on (0, positive infinity)

Judge the parity of function f (x) = (a ^ x + 1) / (a ^ x-1) [a > 0, a is not equal to 1]

f(x)=a^x)^2-1
=a^(2x)-1
exponential function
Non odd non even

If f (x) is a decreasing function on (0, + 00) [0 to positive infinity], and f (a ^ x) is an increasing function on (- 00, + 00) [negative infinity to positive infinity], then the value range of real number a is?
Can you make it clear?

F (x) is a decreasing function on (0, + 00) [0 to positive infinity], while f (a ^ x) is an increasing function on (- 00, + 00) [negative infinity to positive infinity],
So, a ^ x is a decreasing function on (- 00, + 00) [negative infinity to positive infinity],
So, 0