①.y = | x- 1 |-| x-2 |, and the process is accompanied by images.
Analysis: (1) First find the zero value (the root of the equation);
By y = | x- 1 |-| x-2 | = 0
X= 1, or x=2.
The zero value (the root of the equation) is: x= 1, or x=2.
(2) Re-partition discussion:
∫y = | x- 1 |-| x-2 |，
When x
When 1≤x≤2, y = (x-1)-[-(x-2)] = x-1+x-2 = 2x-3;
When x>2, y = (x-1)-(x-2 = x-1-x+2 =1.
Images with different functions corresponding to different intervals are made and combined respectively.
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②. Known: f (1-x/1+x) = (1-x? )/( 1+x？ ), find the analytical formula: f(x)
Method of substitution's thinking is very direct.
"exchange yuan" should examine the definition domain,
f( 1-x/ 1+x)=( 1-x？ )/( 1+x？ The domain of): x ∈ r.
Solution: let t=( 1-x)/( 1+x)
( 1+x)t= 1-x
tx+x= 1-t
x=( 1-t)/( 1+t)
f(t)=( 1-x？ )/( 1+x？ )
=[ 1-( 1-t)？ /( 1+t)？ ]/[ 1+( 1-t)？ /( 1+t)？ ]
=[( 1+t)？ -( 1-t)？ ]/[( 1+t)？ +( 1-t)？ ]
=[( 1+2t+t？ )-( 1-2t+t？ )]/[( 1+2t+t？ )+( 1-2t+t？ )]
=(4t)/(2+2t？ )
= 2t/( 1+t & amp； # 178; )
Change t back to x, and you get it.
f(x)=2x/( 1+x？ )
The domain is still: x ∈ r.
The analytical formula of ∴f(x) is: f(x)=2x/( 1+x? ).
③ It is known that f﹙x﹚ is a linear function, and f [f﹙x﹚] = 4x- 1. Find the analytical formula of f﹙x﹚.
Solution: let f (x) = kx+b.
Then f [f (x)] = k (kx+b)+b.
=k？ x+(kb+b)
=k？ x+b(k+ 1)=4x- 1
So k? =4
b(k+ 1)=- 1
K = 2, substitute b(k+ 1)=- 1 to get b= 1, or b=- 1/3.
The analytical formula of ∴ f x is:
F(x)=-2x+ 1, or f(x)=2x- 1/3.
④ Make piecewise function y = | x- 1 |+| x+2 |,? Use image processing
Analysis: (1) First find the zero value (the root of the equation);
By y = | x- 1 |+| x+2 | = 0
X=-2, or x= 1.
The zero value (the root of the equation) is: x=-2, or x= 1.
(2) Re-partition discussion:
∵y=|x- 1|+|x+2|，？
When x
When -2≤x≤ 1, y =-(x-1)+(x+2) =-x+1+x+2 = 3;
When x> is at 1, y = (x-1)+(x+2) = 2x+1.
Images with different functions corresponding to different intervals are made and combined respectively.
The following figure
⑤. Make the function y = | x? Function image of -2x-3 |.
Analysis: ∫y = | x? -2x-3|=|(x+ 1)(x-3)|
The zero value is: x=- 1, or x=3.
When x
When-1≤x≤3, y=-(x? -2x-3)=-x？ +2x+3
When x>3 o'clock, y=x? -2x-3 .
Images with different functions corresponding to different intervals are made and combined respectively.
You can also do it first.
y=x？ -2x-3 image
Then put y
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