If the maximum value of quadratic function f (x) = x2 + 4x + 6 on [M, 0] is 6 and the minimum value is 2, then the value range of real number m is______ ．

F (x) = x2 + 4x + 6 = (x + 2) 2 + 2, the image is as shown in the figure, let x2 + 4x + 6 = 6, we can get x = - 4 or x = 0 ∵ the maximum value of quadratic function f (x) = x2 + 4x + 6 in [M, 0] is 6, the minimum value is 2, the value range of real number m is [- 4, - 2], so the answer is: [- 4, - 2]

Let ∈ R, then ＂ φ = 0 ＂ is the ()
A. Sufficient and unnecessary condition B. necessary and insufficient condition C. sufficient and necessary condition D. neither sufficient nor necessary condition

Because when φ = 0, f (x) = cos (x + φ) = cosx is even function, it is true; but when f (x) = cos (x + φ) (x ∈ R) is even function, φ = k π, K ∈ Z, it is impossible to deduce φ = 0. So "φ = 0" is a sufficient and unnecessary condition for "f (x) = cos (x + φ) (x ∈ R) to be even function"

Let f (x) be an even function defined on R. when 0 ≤ x ≤ π / 2, f (x) = cos (x π / 3) - 1 / 2, and f (x π) = f (x)
Let f (x) be an even function defined on R. when 0 ≤ x ≤ π / 2, f (x) = cos (x + π / 3) - 1 / 2, and f (x + π) = f (x)
1) Try to write the analytic expression of F (x) on (- π / 2, π / 2)
2) Finding f (31 π / 6)
3) Can we draw the diagram of F (x) in [- 3 π / 2,3 π / 2], and observe and write all the symmetry axes of the function

1) If f (x) is even function on R, then f (x) = - f (- x) is known 0 ≤ x ≤ π / 2, f (x) = cos (x + π / 3) - 1 / 2; then f (x) = - f (- x) = - cos (- x + π / 3) + 1 / 2 = cos (x - π / 3) + 1 / 22) f (31 π / 6) = f (π + 1 π / 6) = f (1 π / 6) = - 1 / 23) is even function with π as periodic Hu function

If f (x) = cos (x + θ) (θ∈ R) is an even function, then a value of θ is?

f(-x)=f(x)
cos(-x+θ)=cos[-(x-θ)]=cos(x-θ)=cos(x+θ)
x=0

Given the function f (x) = a ^ (x + 1) (A & gt; 0 and a ≠ 1), the image passes through the point (0,1 / 2)

Substituting point (0,1 / 2) into f (x) yields
f(0)=a^(0+1)=1/2
The solution is a = 1 / 2
Answer: a = 1 / 2

Given the function f (x) = b * {a} ^ {x} (where a and B are constants and a ＞ 0, a ≠ 1), the image passes through a (1,6), B (3,24)
(1) Find the analytic expression of F (x). (2) if the inequality (A / b) ^ 2 ≥ 2m + 1 holds on X belonging to (negative infinity, 1), find the value range of real number M. calculate f (x) = 3.2 ^ x, find the second question!

(1) Bring in two
ab=6
From these two formulas, we can get a = 2, B = 3
So f (x) = 3 * 2 ^ X
(2)
Another g (x) = (A / b) ^ x = (2 / 3) ^ x
g'(x)=x(ln2-ln3)
It should come to mind later that is to find the derivative of G (x) and then find the minimum value, so that G (x) min > = 2m + 1

If the function f (1 / x) = 1-x / x, what is f (x) = when x is not equal to 0,1

This problem is solved by the method of substitution. Shilling t = 1 / X. get x = 1 / T. get f (x) = 1 / (x-1)

The function f (x) = 1 / X (x) is known

F (x + 1) = (x + 1) 1 / 2 (x + 1)

If f (x) is equal to 2, x times minus 1 / 2 of the absolute value x times of 2

2 ^ X-1 / (2 ^ | x |) = 2, take 2 ^ x as a whole, XX = 2 ^ x, multiply XX on both sides
xx^2-2xx-1=0
Get xx & nbsp; = 1 & nbsp; - & nbsp; sqrt [2] (rounding), & nbsp; XX & nbsp; = 1 & nbsp; + & nbsp; sqrt [2]
x=log（2,1 + Sqrt[2]）

Let f (x) = asin2x + cos2x + a pass through the point (π / 4,2) (1) find the value of a (2) find the monotone increasing interval of the function y = f (x)

2x=π/2,y=2
2=a+0+a
So a = 1
f(x)=sin2x+cos2x=√2(√2/2*sin2x+√2/2cos2x)+1
=√2(sin2xcosπ/4+cos2xsinπ/4)+1
=√2sin(2x+π/4)+1
Incremental is
2kπ-π/2

If the maximum value of quadratic function f (x) = x2 + 4x + 6 on [M, 0] is 6 and the minimum value is 2, then the value range of real number m is______ ．