Author Topic: Additional Math Help HERE ONLY...!  (Read 75468 times)

Offline Sweet_03

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Re: Additional Math Help HERE ONLY...!
« Reply #75 on: June 02, 2009, 02:43:46 pm »
Hello ,

I have a Question .. and I hope You Can Help Me

Paper 1 May-June 2005

Question 6 :
Given that the following functions is defined for the domain -2 ? x ? 3, Find The Range of

(i) f:x ? 2-3x
(ii) g:x ? |2-3x|
(iii) f:x ? 2-|3x|

Ok .. For the First one (i) i know it .. its -7 ? f(x) ? 8
But how does do i work it out for absolute values ?

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #76 on: June 02, 2009, 02:50:31 pm »
Hello ,

I have a Question .. and I hope You Can Help Me

Paper 1 May-June 2005

Question 6 :
Given that the following functions is defined for the domain -2 ? x ? 3, Find The Range of

(i) f:x ? 2-3x
(ii) g:x ? |2-3x|
(iii) f:x ? 2-|3x|

Ok .. For the First one (i) i know it .. its -7 ? f(x) ? 8
But how does do i work it out for absolute values ?
i)2-3*3,2-3*-2 ie  range is -7 to 8
ii)0 to 8 since|2-3x| cannot be -ve but is zero at x=2/3
ii)-7 to 2 sonve x<=2 and is -7 at x=3
« Last Edit: June 02, 2009, 03:02:34 pm by astarmathsandphysics »

Offline sweetsh

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Re: Additional Math Help HERE ONLY...!
« Reply #77 on: June 02, 2009, 02:53:10 pm »
I didnt get the first part by the way 8)

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #78 on: June 02, 2009, 02:59:05 pm »
Paper 2 May june 2004

Question 4

(1 + sec x) (cosec x - cot x) = tan x

Prove the identity. Shoot thats hard.

(1 + sec x) (cosec x - cot x)=(cosx + 1)/cosx* (1 - cosx)/sinx=(1-cos^2x)/(c0sxsinx)=sin^2x/cosxsinx=sinx/cosx=tanx

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #79 on: June 02, 2009, 03:01:26 pm »
sweetsh, did u mean you didnt get what astar just explained?
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Offline sweetsh

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Re: Additional Math Help HERE ONLY...!
« Reply #80 on: June 02, 2009, 03:01:53 pm »
Yes

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #81 on: June 02, 2009, 03:07:01 pm »
Itt is a staright line graph from (3,-7) to(-2,8) so the domain is -2 to 3 and the range is -7 to 8

you read the domain off the x axis and the range off the y axis

Offline moraesikae

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Re: Additional Math Help HERE ONLY...!
« Reply #82 on: June 02, 2009, 03:08:12 pm »
when finding the range with given domain in  quadratic functions,, do you just substitude the values of the given domain into the function??

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #83 on: June 02, 2009, 03:10:26 pm »
be careful with quadratics,
the range only go down to the minimum or up to maximum point

with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain
A genius is 1% intelligence, 99% effort.

Offline moraesikae

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Re: Additional Math Help HERE ONLY...!
« Reply #84 on: June 02, 2009, 03:13:29 pm »
be careful with quadratics,
the range only go down to the minimum or up to maximum point

with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain

ohh thx.. so for quadratics,, you find the minimum point,, and then the maximum point of the curve by substituting to get the largest value of the domain??

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #85 on: June 02, 2009, 03:13:47 pm »
be careful with quadratics,
the range only go down to the minimum or up to maximum point

with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain

No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10

Offline moraesikae

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Re: Additional Math Help HERE ONLY...!
« Reply #86 on: June 02, 2009, 03:16:41 pm »
be careful with quadratics,
the range only go down to the minimum or up to maximum point

with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain

No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10


so if the domain is given,, e.g -2<x<4??


what would the range be>?

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #87 on: June 02, 2009, 03:24:59 pm »
HOW do u get y
   3^2y +5(3^y-10)=0

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #88 on: June 02, 2009, 03:38:01 pm »
be careful with quadratics,
the range only go down to the minimum or up to maximum point

with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain

No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10


so if the domain is given,, e.g -2<x<4??


what would the range be>?

In the example I gave the minimum of the graph was at x=-3. This is included in my domain but not the one you give -2<x<4. In this case you would sub x=-2 and x=4 to get the range. In the case -4<x<4 x=-3 is included so the range woul be from -10 to (4+3)^2-10=39

Offline moraesikae

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Re: Additional Math Help HERE ONLY...!
« Reply #89 on: June 02, 2009, 03:41:44 pm »
be careful with quadratics,
the range only go down to the minimum or up to maximum point

with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain

No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10


so if the domain is given,, e.g -2<x<4??


what would the range be>?

In the example I gave the minimum of the graph was at x=-3. This is included in my domain but not the one you give -2<x<4. In this case you would sub x=-2 and x=4 to get the range. In the case -4<x<4 x=-3 is included so the range woul be from -10 to (4+3)^2-10=39
o rightttt!!! thx v much!!!!!!!