High school math questions. Ask for answers. . (It shouldn’t be difficult,)

f(3x+1)=4x+3,

Let x=0

Then f(1)=3

Let x=2*sina

y=x/2+√(4-x?)

=sina+2cosa

=root 5*sin (a+Φ), (cosΦ=1/square root 5)

Value range: - square root 5<=y<=square root 5

f (x+6) + f(x)=f(x(x+6))

2f(4)=f(4)+f(4)=f(16)

So f (x^2+6x)

Because f(x) is an increasing function on (0, positive infinity)

So x^2+6x<16

Solution-8

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5 answer time: 2009-9-20 17:07 | Let me comment

Ask TA for help Respondent: quater484 | Level 2

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The person downstairs is careless. , points are deducted, 4*4=16, one point for overwhelming a row of people, one point is worth tens of thousands.

Thank you both for writing so many difficult mathematical symbols.

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When X+6>0,X>0, f(x+6)+f(x)=f(x^2+6x)< 2f(4)=f(4)+f(4)=f(16), and because f(x) is an increasing function when X>0, so x^2+6x<16, so 0

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Suppose a linear function f(x)=kx+b

f[ f(x)]=k(kx+b)+b

=k?x+kb+b

=4x+6

Then k? =4 and kb+b=6

The solution is k=±2

①When k=2, b=2

②When k=-2 When b=-6

The analytical formula of ∴f(x) is f(x)=2x+2 or f(x)=-2x-6

After sorting, we get y =x?-4x+6=x?-4x+4+2=(x-2)?+2

The axis of symmetry of the function is x=2, the opening is upward, and in X∈[1,5 ), when x=2, the minimum value is 2, and when x=5, the maximum value is 11. Because x=5 is an open interval, the value range is [2,11). (When the graph of the quadratic function opens upward, the function obtains the minimum value at the symmetry axis point, and the further away from the symmetry axis, the greater the function value; when the quadratic function graph opens downward, the function obtains the maximum value at the symmetry axis point, and the distance from the symmetry axis The farther the axis is, the smaller the value of the function is. Share it with your friends: iTieba, Sina Weibo, Tencent Weibo, QQ Space, Renren, Douban, MSN

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