(1) proves that f(x) is increasing function,
(2) Find the maximum and minimum value of f(x).
② Let the function f(x)=ax? +1\bx+c is odd function (a, b, c∈Z), f (1) = 2, f (2) < 3. Find the values of a, b and C.
③ Let the quadratic function f(x)=x? -4x- 1 Find the maximum and minimum values of the function in the interval.
(1) Proof:
Let x 1, x2∈R, x 1 < x2.
F(x+y)=f(x)+f(y)- 1, namely f(x+y)-f(x)=f(y)- 1.
f(x2)-f(x 1)= f(x2-x 1)- 1
∫x 1 < x2
∴x2-x 1>0
According to the conditions, f (x2-x 1) > 1.
∴f(x2-x 1)- 1>0
That is, f (x2)-f (x 1) > 0.
The function is an increasing function.
(2)
∵ The function is increasing function (proved).
When x= 1, f(x) takes the minimum value; When x=2, f(x) takes the maximum value.
Split 3 into 1+2, and then split 2 into 1+ 1.
f(3)= f(2)+f( 1)- 1 = f( 1)+f( 1)- 1+f( 1)- 1 = 3f( 1)-2 = 4
F( 1)=2。
f(2)=2f( 1)- 1=3
The maximum value of ∴f(x) is 3 and the minimum value is 2.
②
The function ∫ is odd function, and f( 1)=(a+ 1)/b+c=2.
∴f(- 1)=(a+ 1)/-b+c=-2
The numerator is the same, and the result is reciprocal, so the denominator is reciprocal.
The solution is c=0.
F( 1)=(a+ 1)/b=2, that is, a+ 1=2b.
f(2)=(4a+ 1)/2b0
X 1，x2∈。
∴x 1x2+x 1+x2+ 1>0
That is, f (x2)-f (x 1) > 0.
The function is an increasing function.
∴ When x= 1, the function obtains the minimum value f( 1)=0.
When x=3, the function gets the maximum value f(3)= 1/2.
To sum up, the maximum value of this function is 1/2, and the minimum value is 0.