• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Minato112

Minato112

Messages
6,232
Reaction score
5,003
Points
573
bamteck said:
(i) (ii) (iv)
Click to expand...

(i) : gf(x) --> U take f(x) and then U replace in g(x)
f(x) = 2x + 1
g(x) = (2x - 1)/ (x +3)

gf(x) = (2(2x+1) + 1) / ((2x + 1) +3)
= (4x + 3) / (2x + 4)

Now, gf(x) = x
that is, (4x +3) / (2x +4) = x
4x + 3 = 2x^2 + 4x
2x^2 - 3 = 0
x ^ 2 = (3/2)
x = *root* (3/2)

(ii) : To find f inverse, U let the function f(x) to be equal to y

y = f(x)
y = 2x + 1
Now make x the subject of formula

x = (y -1)/ 2
Therefore f inverse : (x - 1)/2 (U replace the y and make it become x)

For g(x)

y = g(x)
(2x - 1)/ (x +3) = y
2x -1 = xy + 3y
2x - xy = 3y + 1
(2 - y)x = 3y + 1
x = (3y +1)/(2 - y)

Therefore g inverse : (3x +1) / (2 - x)

(iii) To sketch the graphs, first U sketch the graph of f(x).. (I believe this one is quite easy)

Now for the inverse, all that we have to do is to reflect the graph in the line y = x.

The reflected part will show the graph of f inverse relative to f(x).

Note : Any calculations were done mentally so am not sure if the answers are correct but I've explained all my workings... Hope it helps.
 
Minato112

Minato112

Messages
6,232
Reaction score
5,003
Points
573
Soldier313 said:
yea it's right!!!:) thanx!
Click to expand...


I'll use "{" as the integration sign.. srry I couldnt find any better

= { 8 sin ((1/2)x) - tan ((1/2)x) dx

= { 8 sin ((1/2)x) dx - { tan ((1/2)x) dx

= 8 { sin ((1/2)x) dx - { tan ((1/2)x) dx

= 8 * 2 * - cos ((1/2)x) - { [sin ((1/2)x)/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) - (-1/2) [ ((-1/2) (sin ((1/2)x))/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) + (1/2) ln |cos ((1/2)x)| + c

Srry it appears bulky but hope U understand.
 
Soldier313

Soldier313

Messages
958
Reaction score
3,499
Points
253
minato112 said:
= -16 cos ((1/2)x) - (-1/2) [ ((-1/2) (sin ((1/2)x))/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) + (1/2) ln |cos ((1/2)x)| + c

Srry it appears bulky but hope U understand.
Click to expand...

JazakAllah khair....thanx a ton! but this is where i didn't understand:(
 
Soldier313

Soldier313

Messages
958
Reaction score
3,499
Points
253
minato112 said:
Mention not ;)

Well I corrected it... The dot meant multiply :)
Click to expand...
um well i understood that actually, it's the second part after the minus sign that i didn't quite understand:/

minato112 said:
= -16 cos ((1/2)x) - (-1/2) [ ((-1/2) (sin ((1/2)x))/ (cos ((1/2)x)] dx

= -16 cos ((1/2)x) + (1/2) ln |cos ((1/2)x)| + c
Click to expand...

the parts in purple are the ones i didn't understand:( sorry and thanx....
 
Minato112

Minato112

Messages
6,232
Reaction score
5,003
Points
573
Soldier313 said:
um well i understood that actually, it's the second part after the minus sign that i didn't quite understand:/

Oh k

the parts in purple are the ones i didn't understand:( sorry and thanx....
Click to expand...

U don't know this rule :

1) { (f ' (x))/ (f (x)) dx = ln | f(x) | + c

2) { 1 / (ax + b) dx = (1/a) { a/ (ax + b) dx
= (1 /a) ln | ax +b | + c
 
Soldier313

Soldier313

Messages
958
Reaction score
3,499
Points
253
minato112 said:
U don't know this rule :

1) { (f ' (x))/ (f (x)) dx = ln | f(x) | + c

2) { 1 / (ax + b) dx = (1/a) { a/ (ax + b) dx
= (1 /a) ln | ax +b | + c
Click to expand...
nop i didn't know those rules! i understood it all now!:):)
JazakAllah khair sooooo much! May Allah bless you inshaAllah. :)
 
whitecorp

whitecorp

Messages
164
Reaction score
111
Points
43
redrecession said:
can u tell me how i can use the quadratic equation to find turning point of a curve? Thx
Click to expand...

You will have to complete the square such that it is in the form y= a(x-b)^2 +c.
Then if a>0, the minimum point is given by (b,c) ; if a<0, the maximum point is given by (b,c) .
Alternatively, you can use calculus, ie find dy/dx and set this derivative =0 to obtain the coordinates of the turning point.
Hope this helps. Peace.
 
bamteck

bamteck

Messages
389
Reaction score
202
Points
53
minato112 said:
(i) : gf(x) --> U take f(x) and then U replace in g(x)
f(x) = 2x + 1
g(x) = (2x - 1)/ (x +3)

gf(x) = (2(2x+1) + 1) / ((2x + 1) +3)
= (4x + 3) / (2x + 4)

Now, gf(x) = x
that is, (4x +3) / (2x +4) = x
4x + 3 = 2x^2 + 4x
2x^2 - 3 = 0
x ^ 2 = (3/2)
x = *root* (3/2)

(ii) : To find f inverse, U let the function f(x) to be equal to y

y = f(x)
y = 2x + 1
Now make x the subject of formula

x = (y -1)/ 2
Therefore f inverse : (x - 1)/2 (U replace the y and make it become x)

For g(x)

y = g(x)
(2x - 1)/ (x +3) = y
2x -1 = xy + 3y
2x - xy = 3y + 1
(2 - y)x = 3y + 1
x = (3y +1)/(2 - y)

Therefore g inverse : (3x +1) / (2 - x)

(iii) To sketch the graphs, first U sketch the graph of f(x).. (I believe this one is quite easy)

Now for the inverse, all that we have to do is to reflect the graph in the line y = x.

The reflected part will show the graph of f inverse relative to f(x).

Note : Any calculations were done mentally so am not sure if the answers are correct but I've explained all my workings... Hope it helps.
Click to expand...

Waaaaoooww ! Mentally ? You are good !
Thanks :)
 
You must log in or register to reply here.
Top