20 points for simple function questions, 20 points for the best

Arbitrary x10 x1f(x2) so subtract the function.

Question 4 Arbitrary x10 x2-x1>0 So f(x1)-f(x2)>0

i.e. f(x1)>f(x2) subtraction function.

f(x)=x^2+1,x∈(0,+∞

f(x)=x^2,x∈(0,+∞

Let a0)f(b)-f(a)=b 2-a 2=(a+c) 2-a 2=c 2+2ac>0

So y=x 2 is an increment on (0,+.

On (0,+ is the increment function.

f(x)=x^2,x

Let a|a|>|b| a^2>b^2

f(b)-f(a)=b^2-a^2〈0

So y=x 2 is a subtraction function on (- 0).

is a subtraction function on (- 0).

f(x)=3/x,x∈(-0)

Let the numerator of a be the same, the smaller the denominator, the larger the fraction.

f(b)-f(a)=3/b-3/a〈0

So on (- 0) is a subtraction function.

According to the image, the first one is an increasing function, and the second one is the first downward translation by one unit, which is still an increasing function.

The third is a subtraction function, and the fourth is a subtraction function.

Function images are not to be forgotten.

Sketching, combining numbers, adding, increasing, subtracting, subtracting.

Select a, using the criterion, f(-x)=-f(x).

So it's an odd function.

y=2|cosx|, cosx|The minimum positive period for is . Paint knows that the function from 2 to on increases monotonically.

y=cos(x 2) The minimum positive period is 4Functions from 2 to on are monotonically reduced.

y=tan(-x) =tanx, the minimum positive period is , and the function from 2 to on is monotonically decreasing.

1, f(2)=f(1)+f(2)=1

f(x) increment function.

f(2) is greater than f(1).

So f(1)=0

2. f(3)+f(4-8x)>2

f(4)=f(2)+f(2)=2

f(3)+f(4-8x)>f(4)

I'll know it myself later, give me extra points.

The length of the upper base of the trapezoid is pc=4-x, the length of the lower bottom is ab=4, and the height is cb=4, so the area s=(upper bottom + lower bottom) x height 2=(4-x+4) x 4 2s = 16 - 2x.

The value of x can be 0 x 4