Y = cos | x | period is? Y = | cosx | period is emergency

Y = cos | x | period is? Y = | cosx | period is emergency

The image of y = cos | x | is exactly the same as that of y = cos X & nbsp; & nbsp; t = 2 π
The image of y = | cosx | is that the image of y = cosx is folded below the X axis to the top of the Z axis, and the top does not move
T=π
Look at the picture
If the function f (x) defined on the interval (- 1.1) satisfies 2F (x) - f (- x) = LG (x + 1), then what is the analytic expression of F (x)?
2f(x)-f(-x)=Ig(x+1) ①
① If x of is replaced by - x, we get:
2f(-x)-f(x)=Ig(-x+1) ②
① X 2 + 2
3f(x)=[2lg(x+1)+lg(-x+1)]
f(x)=[2lg(x+1)+lg(-x+1)]/3
If the function f (x) defined on the interval (- 1.1) satisfies 2F (x) - f (- x) = LG (x + 1), then the analytic solution of F (x) is obtained
From 2F (x) - f (- x) = LG (x + 1)...... 1
The definition interval is (- 1,1)
Substitute - X. Yes.
2f(-x)-f(x) = lg(-x+1) ...........2
Formula 1 * 2 + 2 can be obtained
4f (x) - 2F (- x) + 2F (-... Expansion
If the function f (x) defined on the interval (- 1.1) satisfies 2F (x) - f (- x) = LG (x + 1), then the analytic solution of F (x) is obtained
From 2F (x) - f (- x) = LG (x + 1)...... 1
The definition interval is (- 1,1)
Substitute - X. Yes.
2f(-x)-f(x) = lg(-x+1) ...........2
Formula 1 * 2 + 2 can be obtained
4f(x) - 2f(-x) +2f(-x) - f(x) = 2lg(x+1) + lg(-x+1)
Finally, 3f (x) = 2lg (x + 1) + LG (- x + 1)
That is, f (x) = (2lg (x + 1) + LG (- x + 1)) / 3
How do function calculators calculate logarithmic functions? For example, log3 ^ 5, 3 is the base, 5 is the exponent
Most functional calculators only calculate natural logarithms and common logarithms (i.e. the base is e and 10). If you want to calculate logarithms of any base, you have to use the formula for changing base. If you want to calculate log3 (5) [5 is true], you can
log3(5)=ln5/ln3=1.464973521...
Or log3 (5) = log5 / log3 = 1.464873521
If the function f (x) defined on the interval (- 1,1) satisfies 2F (x) - f (- x) = LG (x + 1), then the analytic expression of F (x) is?
F (x) = LG under the third radical [(x + 1) ^ 2 (1-x)]
It should be right. You can try it
It's an analytical expression of the question
Given that the function f (x) = log base is a, the real number is (x ^ 2-ax + 3), (a > 0, a ≠ 1) satisfies that for any real number x1, X2, when X1 < x2 ≤ A / 2, there is always
Given that the function f (x) = log base is a, the real number is (x ^ 2-ax + 3), (a > 0, a ≠ 1) satisfies that for any real number x1, X2, when X1 < x2 ≤ A / 2, there is always f (x1) - f (x2) > 0. Try to find the value range of real number a
The true number of the function f (x) = log (a) [x & # 178; - ax + 3] is m = x & # 178; - ax + 3, which is a quadratic function, and its symmetry axis is x = A / 2. It is also known that when X1 > > m (x1) > m (x2) = = = > > > F (x1) > F (x2), so the base number of the logarithmic function satisfies: a > 1. In addition, for the logarithmic function, the true number must be greater than 0
If the function f (x) defined on the interval satisfies 2F (x) - f (- x) = LG (x + 1), then the analytic expression is
How to solve the problem
In the above equation, another x = - X
Then f (x) and f (- x) are treated as two unknowns and solved
First of all, the definition field is (- 1,1)
then
4f(x)-2f(-x)=2lg(x+1)
2f(-x)-f(x)=lg(-x+1)
Add it up and you'll get
3f(x)=2lg(x+1)+lg(-x+1)
It's a good idea to move 3 over
f(x)=…… (write domain)
That's it
Ask me if you don't understand
LT
It is known that f (x) = 1 + log takes 2 as the base, X as the true number (1 ＜ = x ＜ = 4), and the function g (x) = [f (x)] ＾ 2 + F (x ＾ 2)
1. Domain of function g (x)
2. The range of function g (x)
g（x）=(1＋log2 x)^2+1＋log2 x^2
=(3/2＋log2 x)^2-1/4
Because 1 ≤ x ≤ 4
So 0 ≤ log2 x ≤ 2
Domain [1,4]
Drawing, range [2,12]
If the function f (x) = LG (x2-ax-3) is a decreasing function on (- ∞, - 1 & nbsp;), then the value range of a is______ .
From the meaning of the question, we can get that y = x2-ax-3 is a decreasing function on (- ∞, - 1 & nbsp;) and Y > 0. So there is A2 > 1 and 1 + A-3 ≥ 0. The solution is a ≥ 2, so the value range of a is [2, + ∞), so the answer is: [2, + ∞)
The 2 + base of 3 is 3, and the true number is the power of logarithm function of 5
Calculation: logarithm function with base 3 and true number one sixth + 2 (logarithm function with base 3 and true number root 2) + 2 of 3 + logarithm function power with base 3 and true number five
Log3 (1 / 6) + 2log3 radical 2 + 3 ^ (2 + log35)
=-log3(2*3)+loh32+(3^2)*3^log35
=-log32-1+log32+9*5
=44
Log3 (1 / 6) + 2 (log3 (root 2)) + 3 ^ (2 + log3 (5))
=log3(1/6)+log3(2)+3^2 *3^(log3(5))
=log3(1/6 *2)+9*5
=-1+45
=44
Two of three, what do you mean.
Logarithm function with base 3 and true number 1 / 6 + 2 (logarithm function with base 3 and true number root 2) = - 1
Let the 2 + base of three be three and the logarithm power of the true number be five = X
Then: logarithm function with base 3 and true number x = logarithm function with base 3 and true number 45. X=45，
Then the original formula = - 1 + 45 = 44
F (x) = LG ((2 / 1-x) + a) is an odd function such that f (x)
LG or ln?