1.given the function f(x) = x*2 -4x -5,find the values of:a) f(k-2) b) f(x*2) - x*2 f(1)2.given two functions f(x) = x*2 -3 and g(x) = 3x-4,find the values of:a) f(k+1) -g(k-1) b) f(g(x)) c) g(f(x))3.let f(x) = 1/1-x,finda) f(f(x)) b) f(f(f(x))) c) f

1.given the function f(x) = x*2 -4x -5,find the values of:a) f(k-2) b) f(x*2) - x*2 f(1)2.given two functions f(x) = x*2 -3 and g(x) = 3x-4,find the values of:a) f(k+1) -g(k-1) b) f(g(x)) c) g(f(x))3.let f(x) = 1/1-x,finda) f(f(x)) b) f(f(f(x))) c) f
1.given the function f(x) = x*2 -4x -5,find the values of:
a) f(k-2) b) f(x*2) - x*2 f(1)
2.given two functions f(x) = x*2 -3 and g(x) = 3x-4,find the values of:
a) f(k+1) -g(k-1) b) f(g(x)) c) g(f(x))
3.let f(x) = 1/1-x,find
a) f(f(x)) b) f(f(f(x))) c) f( 1/f(x) ) d) f (1/f(f(x)) )
感激不尽!

1.given the function f(x) = x*2 -4x -5,find the values of:a) f(k-2) b) f(x*2) - x*2 f(1)2.given two functions f(x) = x*2 -3 and g(x) = 3x-4,find the values of:a) f(k+1) -g(k-1) b) f(g(x)) c) g(f(x))3.let f(x) = 1/1-x,finda) f(f(x)) b) f(f(f(x))) c) f
1.a) f(k-2)=(k-2)²-4(k-2)-5=k²-8k-9
b) f(x²)-x²f(1)=(x²)²-4x²-5-x²(1²-4-5)=x^4-4x²-5+8x²=x^4+4x²-5
2.a) f(k+1)-g(k-1)=[(k+1)²-3]-[3(k-1)-4]=k²+2k-2-3k+7=k²-k+5
b) f(g(x))=(3x-4)²-3=9x²-24x+13
c) g(f(x))=3(x²-3)-4=3x²-13
3.a) f(f(x))=1/[1-(1/(1-x))]=(1-x)/(1-x-1)=(x-1)/x
b) f(f(f(x)))=1/[1-((x-1)/x)]=x/[x-(x-1)]=x
c) f(1/f(x))=f(1-x)=1/[1-(1-x)]=1/x
d) f(1/f(f(x)))=f(x/(x-1))=1/[1-x/(x-1)]=(x-1)/(x-1-x)=1-x

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1.a) f(k-2)=(k-2)^2-4(k-2)-5=k^2-4k+4-4k+8-5=k^2-8k+7
b) f(x*2) - x*2 f(1)=(x^2)^2-4x^2-5-x^2(1-4-5)=x^4+4x^2-5
2.a) f(k+1) -g(k-1) =(k+1)^2-3-3(k-1)+4=k^2+2k+1-3-3k+3+4=k^2-k+5
b) ...

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1.a) f(k-2)=(k-2)^2-4(k-2)-5=k^2-4k+4-4k+8-5=k^2-8k+7
b) f(x*2) - x*2 f(1)=(x^2)^2-4x^2-5-x^2(1-4-5)=x^4+4x^2-5
2.a) f(k+1) -g(k-1) =(k+1)^2-3-3(k-1)+4=k^2+2k+1-3-3k+3+4=k^2-k+5
b) f(g(x)) =f(3x-4)=(3x-4)^2-3=9x^2-24x+16-3=9x^2-24x+13
c) g(f(x))=g(x^2-3)=3(x^2-3)-4=3x^2-13
3.a) f(f(x)) =f(1/1-x)=1/1-(1/1-x)=1-1/x
b) f(f(f(x))) =f(1-1/x)=1/1-(1-1/x)=x
c)f( 1/f(x) )=f(1-x)=1/1-(1-x)=1/x
d) f (1/f(f(x)) =f(1/(1-1/x))=1/1-(x/x-1)=1-x

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1.
a. f(k-2)=(k-2)*2-4(k-2)-5=k*2-8k+7
b. f(x*2)-x*2f(1)=x*4-4x*2-5-x*2(1-4-5)=x*4+4x*2-5
2.
a. f(k+1)-g(k-1)=(k+1)*2-3-3(k-1)+4=k*2-k+5
b. f(g(x))=(3x-4)*2-3=9x*2-24x+...

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1.
a. f(k-2)=(k-2)*2-4(k-2)-5=k*2-8k+7
b. f(x*2)-x*2f(1)=x*4-4x*2-5-x*2(1-4-5)=x*4+4x*2-5
2.
a. f(k+1)-g(k-1)=(k+1)*2-3-3(k-1)+4=k*2-k+5
b. f(g(x))=(3x-4)*2-3=9x*2-24x+13
c. g(f(x))=3(x*2-3)-4=3x*2-13
3.
a. f(f(x))=1/[1-1/(1-x)]=(x-1)/x
b. f(f(f(x)))=1/(1-f(f(x)))=x
c. f(1/f(x))=1/(1-(1-x))=1/x
d. f(1/f(f(x)))=1/(1-x/(x-1))=1-x
帮你对比了一下，这个肯定没错了

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1
（a）f(k-2)=(k-2)^2-4(k-2)-5=k^2-8k+7
(b)f(x*2) - x*2 f(1)=(x^2)^2-4*x^2-5-x^2*(1-4-5)=x^4+4x^2-5
2
(a) f(k+1) -g(k-1)=(k+1)*2 -3-[3*(k-1)-4]=k^2-k+5
(b)f(g(x))=(3x-4)^2-3=9x^2-24x...

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1
（a）f(k-2)=(k-2)^2-4(k-2)-5=k^2-8k+7
(b)f(x*2) - x*2 f(1)=(x^2)^2-4*x^2-5-x^2*(1-4-5)=x^4+4x^2-5
2
(a) f(k+1) -g(k-1)=(k+1)*2 -3-[3*(k-1)-4]=k^2-k+5
(b)f(g(x))=(3x-4)^2-3=9x^2-24x+13
(c) g(f(x))=3(x^2-3)-4=3x^2-13
3
a) f(f(x)) =1/(1-(1/(1-x))) =(x-1)/x
b) f(f(f(x)))=1/(1-((x-1)/x))=x
c) f( 1/f(x) )=1/(1-(1-x))=1/x
d) f (1/f(f(x)) )=f(x/(x-1))=1/(1-(x/(x-1)))=1-x （利用a中的结果）

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这些题的意思大同小异，就是说给你一个函数，那么求以下函数的表达式。
1就直接代入就行 f(k-2) =k^2-8k+7; f(x*2) - x*2 f(1)=x^4+4x^2-5
2 f(k+1) -g(k-1)=k^2-k+5;
f(g(x))=9x^2-24x+13
g(f(x))=3x^2-13
3 f(f(x))=1-1/x
剩下的你自己代入吧

1、f(k-2)=(k-2)^2-4(k-2)-5=k^2-4k+4-4k+8-5=k^2-8k+7
f(x*2) - x*2 f(1)=(x^2)^2-4(x^2)-5-x^2(1-4-5)=x^4-4x^2-5+8x^2=x^4+4x^2-5
乘方用^不用*
2、f(k+1) -g(k-1)=(k+1)^2-3-3(k-1)+4=k^2+2k+1-3-3k+3+4=k^...

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1、f(k-2)=(k-2)^2-4(k-2)-5=k^2-4k+4-4k+8-5=k^2-8k+7
f(x*2) - x*2 f(1)=(x^2)^2-4(x^2)-5-x^2(1-4-5)=x^4-4x^2-5+8x^2=x^4+4x^2-5
乘方用^不用*
2、f(k+1) -g(k-1)=(k+1)^2-3-3(k-1)+4=k^2+2k+1-3-3k+3+4=k^2-k+5
f(g(x)) =(3x-4)^2-3=9x^2-24x+16-3=9x^2-24x+13
g(f(x))=3(x^2-3)-4=3x^2-9-4=3x^2-13
3、 f(f(x))=1/{1-[1-1/(1-x)]}=1/{1-[-x/(1-x)]}=1/[1+x/(1-x)]=1/[1/(1-x)]=1-x
f(f(f(x))) =1/[1-(1-x)]=1/x
f( 1/f(x) )=f(1-x)=1/[1-(1-x)]=1/x
f (1/f(f(x)) )=1/[1-1/(1-x)]=1/[-x/(1-x)]=-(1-x)/x=(x-1)/x
为什么你问问题，我解题，而不是我问问题你解题呢？
关键就在于，我写式子，很讲究，一丝不苟，你而总是忽略细节
当然，这也是因为你太聪明我太蠢的缘故！
这么多答案，这么乱，希望你能仔细看，找出没有错误的答案，这对你也是一种进步！

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1.a) f(k-2)=(k-2)²-4(k-2)-5=k²-8k+7
b) f(x²)-x²f(1)= [(x²)²-4x²-5]-x²[1²-4*1-5]=x^4-4x²-5+8x²=x^4+4x²-5
2.a) f(k+1) -g(k-1)= ...

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1.a) f(k-2)=(k-2)²-4(k-2)-5=k²-8k+7
b) f(x²)-x²f(1)= [(x²)²-4x²-5]-x²[1²-4*1-5]=x^4-4x²-5+8x²=x^4+4x²-5
2.a) f(k+1) -g(k-1)= [(k+1)²-3]-[3(k-1)-4]=k²-k+7
b) f(g(x))=f(3x-4)=(3x-4)²-3=9x²-24x+13
c) g(f(x))=g(x²-3)=3(x²-3)-4=3x²-13
3.a) f(f(x))=f(1/(1-x))=1/[1-1/(1-x)]
b) f(f(f(x)))=f(f(1/(1-x)))=f(1/[1-1/(1-x)])=1/{1-1/[1-1/(1-x)]}
c) f( 1/f(x) )=f(1-x)=1/(1-(1-x))=1/x
d) f (1/f(f(x)) )=f (1/f(1/(1-x)) )=f(1-1/(1-x))=1/{1-[1-1/(1-x)]}=1-x

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1. instead the variable then
a) f(k-2)=(k-2)²-4(k-2)-5=k²-8k-9
set f(k-2)=0 factorization reach f(k-2)=(k+1)(k-9)=0 then reach x=-1 or x=9
b) f(x²)-x²f(1)=(x&...

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1. instead the variable then
a) f(k-2)=(k-2)²-4(k-2)-5=k²-8k-9
set f(k-2)=0 factorization reach f(k-2)=(k+1)(k-9)=0 then reach x=-1 or x=9
b) f(x²)-x²f(1)=(x²)²-4x²-5-x²(1²-4-5)=x^4-4x²-5+8x²=x^4+4x²-5
set f(x²)-x²f(1)=0 factorization reach f(x²)-x²f(1)=(x^2+5)(x^2-1) then reach x^2=-5 rejection or x^2=1 then x=-1 or x=1
2. instead the variable then
a) f(k+1)-g(k-1)=[(k+1)²-3]-[3(k-1)-4]=k²+2k-2-3k+7=k²-k+5
set f(k+1)-g(k-1)=0 factorization reach using the formula :
Δ=b^2-4ac≥0，x=[-b±(b^2-4ac)^(1/2)]/2a
Δ=b^2-4ac＜0，x=[-b±(4ac-b^2)^(1/2)]/2a
b) f(g(x))=(3x-4)²-3=9x²-24x+13
same to 2.(a)
c) g(f(x))=3(x²-3)-4=3x²-13 x= radical sign 39/3 or x= - radical sign 39/3
3. instead the variable then
a) f(f(x))=1/[1-(1/(1-x))]=(1-x)/(1-x-1)=(x-1)/x
b) f(f(f(x)))=1/[1-((x-1)/x)]=x/[x-(x-1)]=x
c) f(1/f(x))=f(1-x)=1/[1-(1-x)]=1/x
d) f(1/f(f(x)))=f(x/(x-1))=1/[1-x/(x-1)]=(x-1)/(x-1-x)=1-x

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