How to do these two high school function questions, how to do these high school math function questi

1. If x+2 10-x then x 4, just compare 2 x and 10-x. When x=4, 2 x=16>10-x=6;When x=5, 2 x=32>10-x=5, it can be seen that f(x) is maximum when x=4.

If x+2<10-x then x<4, just compare 2 x with x+2. When x=3, 2 x=8>x+2=5;When x=2, 2 x=4>x+2=4, it can be seen that f(x) is the largest when x=3.

In summary, x=4, f(x) is up to 6. Therefore, C is chosen.

This question is a multiple-choice question, and it is recommended to use the option to bring in the method directly, which is faster.

2. f(x) is an even function on r, and f(x) increases monotonically on [0.

f(|2x-1|) in monotonically increasing on r.

and f(|2x-1|To sum up, I hope it will help you.

I'm sorry, but I'll ask the second question.

Solution: f(x) is an even function on r, f(2x-1) = f(|2x-1|), and f(x) is monotonically increasing on [0, |2x-1|0,1 3>0, and f(2x-1).

In the first question, we should choose c, compare who is greater than whom in the interval of these three numbers, and finally conclude that when x=4, f(x) has a maximum value of 6

The second question is that the original function is an even function, then decreasing in the interval [-infinity, 0], and there is f(-1 3)=f(1 3) then when 2x-1 (-1 3, 1 3), the meaning of the question is satisfied, i.e., x (1 3, 2 3).

The first question [c], this question should be replaced by options, and the substitution of a is f(x)=min[4th power of min[2, 6,6]=6

According to the definition of min[a,b,c] representing the smallest of the three numbers abc, the smallest 6 of the three numbers in this problem is taken as 6, and the following is the same];

If the generation b then f(x) = min[5th power of 2, 7,5] = 5

If c is then f(x) = min[2 to the 6th power, 8,4] = 4;

If a is then f(x) = min[2 to the 7th power, 9,3] = 3

Question 2 [b] This question examines the definition of the even function f(x)=f(-x) f(x)=f(-x)=f(x)=f(丨x丨).

So f(2x-1)=f(丨2x-1丨) f(1 3) 丨2x-1丨 1 3 [because f(x) increases monotonically on the interval [0,+] -1 3 2x-1 1 3 x (1 3, 2 3).

You may guess that the answer to this question is incorrect, and it should not contain 1 3, that is, item b should be (1 3, 2 3) and not [1 3, 2 3).

One. y=f(x+8) is an even function, then the image is shifted to the right by 8 units, and the image becomes an image of f(x), that is, f(x) is symmetrical with respect to x=8. Then according to f(x) in the interval (8,+ is the subtraction function, draw the graph, then choose d

Two. Let the profit be y, the list price is x, and the sales volume is n, and let n=ax+b(a<0) when x=300, n=0, then 300a+b=0

y=(x-100)(ax+b)=a(x-100)(x-300) When x=200, the function achieves the maximum value.

Three. Let x1 and x2 be f(x) to define any two independent variables in the domain, x2 > x1

Then f(x2)-f(x1)=f(x1)*f(x2-x1)-f(x1)=f(x1)[f(x2-x1)-1]>0, so f(x) is an increasing function.

Let x=y=0, then f(0)=f(0)*f(0), f(0)≠0, so f(0)=1;f(negative infinity) = f (negative infinity) * f (negative infinity) > = 0

Then when x<0, f(x) is at (f(negative infinity), f(0)), i.e., 0

This problem has actually been set for you, and the breakthrough point is to prove the monotonicity of functions by algebra and definition.

>0 x>3/2

is the axis of symmetry of the function, the opening of the function is upward, the minimum value is -11 when x=2 is the maximum value is 25 when x=-4

1 all, if and only if x=2 the equal sign holds, so g(x)=x+4 x is the minimum value of 4 on the interval a=[1,5 2], at this time x=2, according to the title, x0=2, f(x0)=g(x0)=4So.

f(x)=(x-2) 2+4 x [1,5 2]When x=1, the value of f(x) on a is the largest, and the maximum value = 5 is selected c2

1) If t<0, take x=t, then x<0, x+t<0f(x+t)=f(2t)=-4x 2

2f(x)=-2x^2

f(x+t)<2f(x), which contradicts the title.

2) If t 0, x 0, x + t 0

f(x+t)=(x+t)^2

f(x)=x^2

Let g(x)=f(x+t)-2f(x)=-x 2+2tx+t 2g(x) take x=t as the axis of symmetry, the opening is downward, and g(x) decreases monotonically on [t,t+2].

To make g(x) 0 constant, only g(t+2) 0 is established, that is, -(t+2) 2+2t(t+2)+t 2 0 is established and organized, and it is obtained: t 2-2 0

t≥0t≥√2

1. [a-1,2a], define the domain with respect to the origin symmetry 2a=1-a to get a=1 3 even function axis of symmetry x=b -2a with respect to y-axis symmetry = 0, and b = 0

2. Drawing [-6,-2] is equivalent to taking 4 as the period of the image [-2,2] so f(x)=(x+4) 2+1

1.Definition.

b=2.If you can draw a picture, you have to draw a picture for this kind of question.

f(x)=x^2+1

2 to the power of a = 5 to the power of b = m

There is a=log2m and b=log5m

1/a+1/b=2

There is 1 log2m+1 log5m=2

log2m+log5m log2m*log5m=2 uses the formula for changing the base

in the calculation