Let f (x) = 1-1 / x + x square judge its parity and prove f (1 / x) = - f (x))

Let f (x) = 1-1 / x + x square judge its parity and prove f (1 / x) = - f (x))

f(x)=(1+x^2)/(1-x^2)
f(-x)=[1+(-x)^2]/[1-(-x)^2]=(1+x^2)/(1-x^2)
So f (x) is even function
f(1/x)=(1+1/x^2)/(1-1/x^2)=(x^2+1)/(x^2-1)=-f(x)

It is known that the function f (x) = x square + BX + C has a unique zero point
It is known that the function f (x) = x square + BX + C has a unique zero point. 1. Find the expression of 1: F (x); 2: the maximum value of F (x) in the interval [a, a + 2] is 4, find a

The function f (x) = x square + BX + C has a unique zero point 1
Namely
b²-4c=0
1+b+c=0
Solution
b=-2
c=1
therefore
1.f(x)=x²-2x+1
two
f(x)=(x-1)²
1)a>=0
Maximum value = f (a + 2) = A & # 178; + 2A + 1 = 4
a²+2a-3=0
(a-1)(a+3)=0
a=1
2) a

If the function f (x) = the square of (a + 1) x + 1 / BX, and f (1) = 3, f (2) = 9 / 2 (1), find the value of a and B, and write the expression of F (x)
(2) It is proved that f (x) is an increasing function on [radical 2 / 2, positive infinity]