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Let a > 0, the x power of F (x) = E / A + the x power of a / E be an odd function. 1. Find a, 2. Prove that it is an increasing function on [0. Positive no group]
Let a > 0, f (x) = the x power of E / A + the x power of a / E be an even function. 1. Find a, 2. Prove that it is an increasing function [solution] on [0. Positive infinity]. F (x) = f (- x) f (x) = e ^ X / A + A / e ^ XF (- x) = e ^ (- x) / A + A / e ^ x = e ^ (- x) / A + A / e ^ (- x) e ^ X / A + A / e ^ x = 1 / (AE ^ x) + AE ^ Xe ^ x (1 / A-A) = 1 / e ^ x (
Given that f (x) = the x-th power of a - the x-th power of a (a > 0 and a is not equal to 1) (1) prove that the function f (x) is an odd function (2) discuss and prove the monotonicity of F (x)
(1)f(x)=a^x-a^(-x)
f(-x)=a^(-x)-a^x=-[a^x-a^(-x)]=-f(x)
The function f (x) is an odd function
(2)
When a > 1, f (x) = a ^ x-a ^ (- x) monotonically increases on R,
When 01, x1
It is known that the function f (x) = 1-A / (x power of 3) + 1 is an odd function. (1) find the value of a; (2) It is proved that f (x) is an increasing function on R; (3) When Party x belongs to [- 2,2), find the value range of function f (x); (4) find the solution set of inequality f [Log1 / 2 (3-x)] + F [1 / 3. Log2 (3-x) - 2 / 3] > = 0. (four questions)
(1) Obviously, if the field of F (x) is r, then f (- x) = 1 - [a * 3 ^ X / (3 ^ x + 1)] because f (x) is an odd function, then f (- x) = - f (x) is 1 - [a * 3 ^ X / (3 ^ x + 1)] = A / (3 ^ x + 1) - 1 is a (3 ^ x + 1) / (3 ^ x + 1) = 2 because 3 ^ x > 0, then a = 2. Another method: because f (x) is an odd function on R, then f (0) = 0, that is, f (0) = 1-A / (3 ^ 0 + 1) = 0, a
Let f (x) be an odd function defined on R. when x is greater than or equal to 0, f (x) = the x power of 2 + 2x + B (B is a constant), then what is f (- 1) = 2
∵ f (x) is an odd function defined on R
∴f(0)=0
I.e. 1 + B = 0
b=-1
F (x) = 2 ^ x + 2x-1 (f (x) meets the test). Don't forget
f(-1)=-f(1)=-[2+2-1]=-3
F (x) = a minus (x power of 2 plus 1 / 2) is an odd function. Find the value of constant a
Another definition can be used;
As an odd function, f (0) = 0;
The 0 power of 2 is 1;
Denominator 1 + 1 = 2;
The molecule is 1;
A-1/2=0;
A = 1 / 2
But this applies to filling in the blanks and choosing. When you encounter big questions
You have to prove again that the domain is symmetrical about the origin;
F (x) = - f (- x), OK
The above method is a panacea
Given that f (x) = 2 divided by (x-1 of 3), plus m is an odd function, find the value of the constant M
f(x)=2/(3^x-1) +m=(2+m3^x-m)/(3^x-1)
-f(x)=(2+m3^x-m)/(1-3^x)
=(m3^x+2-m)/(1-3^x)
f(-x)=2/(3^(-x)-1)+m
=2*3^x/(1-3^x)+m
=(2*3^x+m-m3^x)/(1-3^x)
=((2-m)*3^x+m)/(1-3^x)
2-m=m
m=1
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