function 不会做~~

外星護法

1.） function f is defined by f:x--> x^2+3. Find the function of f^2 and state its range.

2.) the function f and g, each with domain R, are defined by  f:x--> x+1     g:x--> modulus x.
     find the solution set of the equation gf(x)=fg(x)

3.)Given f(x)= (x-2)^2, find the values of x for which f(x)=f^-1(x)...


感激不尽~~

四月一日的小皮

1.f(x) = x^2 +3
  f^2(x) = ff(x)
            = f(x^2 + 3)
            = (x^2 + 3 )^2 + 3
            = x^4 + 6x^2 + 12
when x -> +infinity, f^2(x) -> +infinity
when x -> -infinity, f^2(x) -> +infinity
so when x=0 , f^2(0) = 12
the range for f^2(x) is {f(x):f(x) >= 12}

>= means more or equal

Log

1. f²(x)=ff(x)
          =f(x²+3)
           =x^4+6x²+12,  Range= ﹛y: y ≥12﹜
to get range ,we analyze this x^4+6x²+12 equation.（think）

2. gf(x)=fg(x)
  lx+1l = lxl+1
tips sketch the the two functions you get the solution SET.
3.if u sketch the graph f and f^-1,you will find that f and f^-1 intersect with each other at y=x
that means f(x)=f^-1(x)=x
                 f(x)=x
                   x^2-4x+4=x
solve, get two x values.

四月一日的小皮

3. f^-1(x) = y
         f(x) = y
           y  = (x-2)^2
   (y)^1/2 = x - 2
           x  = (y)^1/2 + 2
thus, f^-1(x) = (x)^1/2 + 2               
         f(x) = f^-1(x)
        (x - 2)^2 =  (x)^1/2 + 2
         x^2 - 4x + 4 =  (x)^1/2 + 2
         x^2 - 4x + 2 =  (x)^1/2
         x^4 - 8x^3 + 20x^2 - 16x + 4 = x
         x^4 - 6x^3 + 20x^2 -17x +4 = 0
.....
then u do urself.
i dun have calculator nw.
correct me if i'm wrong

外星護法

都做到了~~
谢谢~~ thanks~

白羊座aries

3. f^-1(x) = y
         f(x) = y
           y  = (x-2)^2
   (y)^1/2 = x - 2
           x  = (y)^ ...
四月一日的小皮 发表于 11-1-2010 09:52 PM

怎样factorise power of 4 的equation ? calculator也做不到的事情

外星護法

怎样factorise power of 4 的equation ? calculator也做不到的事情
白羊座aries 发表于 12-1-2010 07:07 PM

    trial and error ??

四月一日的小皮

yes,u can use trial and error method.
given that
g(x) =  x^4 - 8x^3 + 20x ....

when x=4, g(4) = 0 ;

when x=1,g(1)= 0.

So u can solve....

x^4 - 8x^3 + 20x^2 -17x +4 = 0
(x - 1)(x - 4)( x^2 - 3x + 1) = 0

then u know that for ( x^2 - 3x + 1) can't factorise.
so,using formula u can find complex number of 1/2 ( 3 -/+ (5)^1/2 ).

till here,we have a problems.
if you draw the curve for both, just as Log say,
it just meet a two points.

as we know for f^-1(x) and f(x) to be trued, one-to-one mapping is the must.

so since, the curves meet at x=1/2 ( 3 - (5)^1/2) and x=4.
Thus,
the answer should be 4 and 1/2 ( 3 - (5)^1/2).


Sry for so long.
I become stupid after coming back from holiday.xD

Log

sorry,请问这题3题有domain for the f(x)= (x-2)^2  ？
if not stated, we take x=4 and 1/2 (3-√5)

注：我刚刚的做法f（x)=x有点mistakes。

walrein_lim88

应该是4个答案吧，若没有RESTRICTION OF DOMAIN。。

四月一日的小皮

thk you,u're right.
answer is 1,4 plus two complex number.


[img]URL=http://img693.imageshack.us/i/20258497.png/][/URL][/img]
here is the pic of two curve.

walrein_lim88

不过：
to be inverse function, the function itself must be one-one function.
By right, there should have restricted domain.
If there have restricted domain where one-one mapping is occured,
then LOG的方法是对的：
f(x)=f^-1(x) => f(x)=(x) because they will only meet at the line y=x.

四月一日的小皮

哈哈，我数学退步了啦。

终结：
Since for inverse function, one-to-one mapping is the must.
So, the answer should be 1 and 4,as they will only meet at the line y=x, just as walrein_lim88 and Log say.

if the question is not function-related,
let say (y)^1/2 = x - 2 is just a equation to be solved.
thus,in here, the 4 answer must be included.

有点不会了。T－T

Log

不过：
to be inverse function, the function itself must be one-one function.
By right, there shoul ...
walrein_lim88 发表于 13-1-2010 10:17 AM

其实f(x)=f^-1(x).
      (x-2)^2= ±√x+2,   ±都可以因为没有restriction of domain

          (x - 1)(x - 4)( x^2 - 3x + 1) = 0
if do in this way,we will get another 2 values that does not satisfy the (1) ,that are, 1 and 1/2 (3+√5).
1 and 1/2 (3+√5) only satisfy  -√x + 2=(x-2)^2
4 and 1/2 (3-√5)        satisfy    √x + 2=(x-2)^2
【至于会拿到another 2 values 是因为（ x^2 - 4x + 2 ）^2= (±x)^2 】

but，这问题有错误，（x-2^2)不是one to one function，so there must have domain for it to be one to one mapping.
if have domain , my method can be used.（我是说method)。

这method要看f and f^-1intersect with each other at y=x才能用 。

如有错误，请指点。

Log

回复 13# 四月一日的小皮
你的答案也是对的，其实我们大家被这absence of domain of  f 给混淆了


还有，学长 1/2 (3±√5)不是叫complex number，应是real number

四月一日的小皮

回复  四月一日的小皮
你的答案也是对的，其实我们大家被这absence of domain of  f 给混淆了


还有， ...
Log 发表于 13-1-2010 06:53 PM

我人老认命了。
我相信你在将来的stpm里math一定考到A的。
加油。

外星護法

paiseh~~ the function f is with domain {x <R, x>= 2} ... 一时没有打到～ sorry~~

and may i ask one more question ?

for no 3, actually how we can know f(x) intersect f^-1(x) at y=x by just sketching the curve ?

谢谢～

外星護法

我人老认命了。
我相信你在将来的stpm里math一定考到A的。
加油。
四月一日的小皮 发表于 13-1-2010 07:35 PM

LOG 是今年STPM考生 ？？

walrein_lim88

paiseh~~ the function f is with domain {x = 2} ... 一时没有打到～ sorry~~

and may i ask one more  ...
外星護法 发表于 13-1-2010 08:13 PM

蓝色是f(x)=(x-2)^2
红色是f^-1(x)
绿色代表y=x (reflection line)
Look at the diagram ,where is the intersection point?
at the line y=x .. (x=4)

for inverse function to be worked, function itself must be:
a)one-one mapping
b)onto function (all object has matched with all image)

since only one-one mapping can be inversed:
look at the curve, after restricted where domain x,>or = 2, den it become one-one mapping.
look at graph, after inverse, it only will meet wif f(x) at the line y=x
therefore, f(x)=f^-1(x) => f(x)=x

Log

回复 18# 外星護法

对，一起加油吧今年的stpm