Let x=0
Then f( 1)=3.
Let x = 2 * Sina.
y=x/2+√(4-x？ )
= Sina +2cosa
= radical number 5 * sin (a+φ), (cosφ= 1/ radical number 5)
Range:-Number of roots 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 increasing function at (0, positive infinity).
So x 2+6x
Solution -8
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Answer time: September 20, 200917: 07 | Let me comment.
Respondents seeking TA help: quater484 | Level 2
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Downstairs, careless, penalty points, 4*4= 16, one point overwhelms a row of people, and one point is tens of thousands.
Thank you 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), because X>0 f(x) is increasing function.
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Let the 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.
(1) when k=2, b=2.
② When k=-2, b=-6.
The analytical formula of ∴f(x) is f(x)=2x+2 or f(x)=-2x-6.
Finishing, y=x? -4x+6=x？ -4x+4+2=(x-2)？ +2
The symmetry axis of the function is x=2, and the opening is upward at X∈[ 1, 5]. The minimum value is 2 when x=2, and the maximum value is 1 1 when x=5. Because x=5 is an open interval, the value range is [2, 16558]. (When the image opening of quadratic function is upward, the function takes the minimum value at the point of symmetry axis, and the farther away from the symmetry axis, the greater the function value; When the image opening of quadratic function is downward, the function gets the maximum value at the symmetry axis, and the farther away from the symmetry axis, the smaller the function value. Share it with your friends: I post Sina Weibo Tencent Weibo QQ Space Everyone Douban MSN.
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