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On Mathematics Problems in Senior One (required 1)
Function representation-image method;

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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