Soalan berkaitan Matematik & Matematik Tambahan

BeanDiesel

kalo kurang jelas sila tanye. toksah memalu.

Robab

Originally posted by BeanDiesel at 26-3-2006 09:39 PM


i'm not familiar with the notation 'f:x'. f:x tu f as a function of x ke? like f(x)?

let's say it's f(x)=-x^2 + 3x + 6

you wanna find the values of x when f(x)=2x

so, -x^2 + 3x + 6 = 2 ...

:clap: thanks. :bgrin:
jadik snang plak kalau ade org tlg tunjukkan...
maseh yekk...
nanti ade lagi
sy tanye lagi...

Robab

g(x) = x - 5, x # 5.

find : g^-1(x)
(slalu berbelit kalau soklan cenggini terutama bile nk convert from y to x)

BeanDiesel

Originally posted by Robab at 3/26/06 06:15 AM
g(x) = x - 5, x # 5.

find : g^-1(x)
(slalu berbelit kalau soklan cenggini terutama bile nk convert from y to x)

lame tak wat benda ni. cuba jap.

g(x) = x - 5

let y = g(x)

so, y=x-5

x=y+5  <---x as a function of y

so, g^-1(y) = y+5

g^-1(x) = x + 5

g^-1(z) = z+5

g^-1(w) = w + 5 bla bla bla....doesn't matter
since the problem wants the inverse of g as a function of x
maka
amik yang ni
g^-1(x) = x + 5




check
g(8) = 3

g^-1(3) = 8  <---kite nak jadi camni

g^-1(x) = x + 5
g^-1(3) = 3 + 5 = 8....ok




well, topic ni kalo susah nak faham aku rasa leh hafal jek steps dia.

say u wanna find the inverse of f(x)

1)set f(x) = y
2) y=bla bla bla x (y as a function of x) ---> x = bla bla bla y (x as a function of y)
3) f^-1(y) = bla bla bla y
4) just tuka jek y dalam step 3 tu x, f^-1(x) = bla bla bla x

inverse ni susah gak nak explain, at least bagi aku la.

Robab

arigato gozaimasu sensei

BeanDiesel

doo itashima****e.

Robab

g(x) =   2   , x #5.
          x - 5

Find:
a) g^1(x)



*yg td tu biase jek
ni ade pecahan plak...
sorilaaa slalu kantoi ngan pecahan ni...

BeanDiesel

Originally posted by Robab at 3/26/06 06:47 AM
g(x) =   2   , x #5.
          x - 5

Find:
a) g^1(x)



*yg td tu biase jek
ni ade pecahan plak...
sorilaaa slalu kantoi ngan pecahan ni...

jom test ikut langkah yang aku kasi tadi

1)set g(x) = y

g(x)= 2/(x-5) = y



2) y=bla bla bla x (y as a function of x) ---> x = bla bla bla y (x as a function of y)

y= 2/(x-5)   <---y as a function of x
x-5 = 2/y
x= 2/y  + 5   <x as a function of y



3) g^-1(y) = bla bla bla y

g^-1(y) =  2/y  + 5



4) just tuka jek y dalam step 3 tu x, g^-1(x) = bla bla bla x

g^-1(x) = 2/x + 5




check

g(x)= 2/(x-5)

g(3) = 2/ (3-5) = -1

adakah g^-1( -1) = 3?

g^-1(-1) = 2/-1 + 5 = 3.....ok

Atomic_Omnikid

diberi dp/dt=kt^2 dgn k=pemalar,diberi juga p=3 2/3 dan dp/dt =-2 apb t= -2

[a)ungkapkan p dlm seb. t
[b)cari nilai p apb t=6

[ Last edited by Atomic_Omnikid at 26-3-2006 11:26 PM ]

Robab

f(x) = x + 3
fg(x) = 5x + 13. determine the function g

version sy buat :

f(x) = x + 3
fg(x) = 5x + 13
f[g(x)] = 5x + 13
x + 3g(x) = 5x + 13
3g(x) = 5x + 13 - x
g(x) = 4x + 13
               3

version jawapan plak :

(x + 3) = 5(x + 3) - 2
y = 5(y) - 2
g(x) = 5x - 2

soalan sy : camane die boleh dpt cenggitu? mane -2 tu datang? :stp:
sy skang tgh refer 2 books
yg satu buku tu slalu kasik jawapan in my version
yg this 2nd book tu plak slalu buat jwpn cenggitu...xtau yg mane satu betul...


nota kaki : sy kene belajar sniri addmath sbb tertinggal 2 chapters dah ngan bebudak lain...sebab tu jadi lembab sket tu...heheh

shada

jawapan sy plak gini...

f(x)= x + 3
fg(x) = 5x + 13

f[g(x)] + 3 = 5x + 13
g(x) = 5x + 13 - 3
g(x) = 5x + 12

******************************************
:hmm::hmm::hmm:

Robab

Originally posted by shada at 30-3-2006 04:06 PM
jawapan sy plak gini...

f(x)= x + 3
fg(x) = 5x + 13

f[g(x)] + 3 = 5x + 13
g(x) = 5x + 13 - 3
g(x) = 5x + 12

******************************************
:hmm::hmm::hmm:

cuba perhatikan yg saya bold tuh

13-3 = 10
5x + 10
erm
nanti kita tunggu reply org yg boleh tlg kita

BeanDiesel

Originally posted by Robab at 3/29/06 01:29 AM
f(x) = x + 3
fg(x) = 5x + 13. determine the function g

version sy buat :

f(x) = x + 3
fg(x) = 5x + 13
f[g(x)] = 5x + 13
x + 3g(x) = 5x + 13
3g(x) = 5x + 13 - x
g(x) = 4x + 13
               3

version jawapan plak :

(x + 3) = 5(x + 3) - 2
y = 5(y) - 2
g(x) = 5x - 2

soalan sy : camane die boleh dpt cenggitu? mane -2 tu datang?
sy skang tgh refer 2 books
yg satu buku tu slalu kasik jawapan in my version
yg this 2nd book tu plak slalu buat jwpn cenggitu...xtau yg mane satu betul...


nota kaki : sy kene belajar sniri addmath sbb tertinggal 2 chapters dah ngan bebudak lain...sebab tu jadi lembab sket tu...heheh

f(x) = x + 3
fg(x) = 5x + 13. determine the function g

fg(y)=5y+13  <----same jek ngan yang atas, just nak kasi clear sket (takyah pon takpe)
now, g(y) ialah argument of f(x), meaning that, g(y) is treated like a variable x of f(x)

maknanya

f(g(y)) = g(y) + 3 = 5y + 13

so,
g(y)= 5y +13 - 3

so g(y) = 5y + 10

g(x) = 5x + 10

shada buat
g(x) = 5x + 13 - 3
g(x) = 5x + 12    <----mistake in subtraction, still akan dapat jawapan yang same ngan aku nye.

jap, aku pikir lagi

[ Last edited by BeanDiesel at 30-3-2006 02:00 AM ]

BeanDiesel

oo, aku tau mane dia dapat g(x) = 5x - 2

f(x) = x + 3

kalo g(f(x)) = 5x + 13, yes, g(x) = 5x - 2

caranya (well, one of the ways)

say  g(y) = Ay + B  where A and B are constants

so,

g( f(x)) = A( f(x) ) + B = A (x+3) + B = 5x + 13

Ax + 3A + B = 5x + 13

so, 5x = Ax, maka A = 5

3A + B = 13
B = 13 - 3A = 13 - 3(5) = 13 - 15 = -2

therefore,
g(y) = Ay + B = 5y - 2
g(x) = 5x - 2

tu kalau g ( f(x) ) = 5x + 13 la,
kalau f ( g(x)) = 5x + 13 aku rasa jawapan aku ngan shada betul.

BeanDiesel

oh yeah, convention yang aku pakai is

f (g(x)) ----> f as a function of g(x), maknanya, g(x) tu argument of f(x)

aphrodisiac

Originally posted by Atomic_Omnikid at 26-3-2006 11:25 PM
diberi dp/dt=kt^2 dgn k=pemalar,diberi juga p=3 2/3 dan dp/dt =-2 apb t= -2

[a)ungkapkan p dlm seb. t
[b)cari nilai p apb t=6

senang je ni...

a)

since dp/dt = -2 when t = -2, so

-2 = k(-2)^2 ,   solving ........k = -1/2

integrating dp/dt = kt^2  yields p = k t^3/3 + C  , C = constant
                                                   p = -t^3/6 + C

given that p = 11/3 when t = -2, so
                                              11/3 = 8/6 + C
                                                  C = 7/3

p(t) = -t^3/6 + 7/3

b) p(6) = -216/6 + 7/3 = -101/3

Robab

Originally posted by BeanDiesel at 30-3-2006 06:12 PM
oh yeah, convention yang aku pakai is

f (g(x)) ----> f as a function of g(x), maknanya, g(x) tu argument of f(x)

thanks BD

Robab

Originally posted by Robab at 29-3-2006 05:29 PM
f(x) = x + 3
fg(x) = 5x + 13. determine the function g

version sy buat :

f(x) = x + 3
fg(x) = 5x + 13
f[g(x)] = 5x + 13
x + 3g(x) = 5x + 13
3g(x) = 5x + 13 - x
g(x) = 4x + 13
   ...

sy salah kt situlah kn
sepatutnya
g(x) + 3 = 5x + 13
g(x) = 5x + 10
:bgrin:
baru paham care nk pindah die tuh
heh

shada

Originally posted by Robab at 30-3-2006 05:27 PM

cuba perhatikan yg saya bold tuh

13-3 = 10
5x + 10
erm
nanti kita tunggu reply org yg boleh tlg kita

:kantOoppsssss!!!!!!!!!!!!! salah laaa plak! HARAP MAAF! MASALAH TEKNIKAL YANG TIDAK DIDUGA.

Robab

yer
biase ah tuh
ngeh ngeh