O-Level Additional Math

[Augmum]

LOL123:

pls help me


The function f is defined by f(x)=1-2x.
(a) Find the range of f corresponding to the domain -1≤x<3
(b) Find the domain of f corresponding to the range 0≤f(x)<5

thanks

Post on behalf on my gal....

(a) subst x = -1 into the equation
y = 1 - 2(-1)
y = 3
Substx = 3 into the equation
y = 1 - 2(3)
y = -5

Thus, the range of f is : { y: -5 < y ≤ 3}

(b) qn means that: find the range of values of x such that f(x) = 0
1 -2x = 0
2x = 1
x = 0.5

1 - 2x = 5
2 x = -4
x = - 2

{x : -2 < x ≤ 0.5 }

Note: The domain of the function refers to the x value while the range of the function refers to the y value.

[LOL123]

Augmum:

LOL123:

pls help me


The function f is defined by f(x)=1-2x.
(a) Find the range of f corresponding to the domain -1≤x<3
(b) Find the domain of f corresponding to the range 0≤f(x)<5

thanks

Post on behalf on my gal....

(a) subst x = -1 into the equation
y = 1 - 2(-1)
y = 3
Substx = 3 into the equation
y = 1 - 2(3)
y = -5

Thus, the range of f is : { y: -5 < y ≤ 3}

(b) qn means that: find the range of values of x such that f(x) = 0
1 -2x = 0
2x = 1
x = 0.5

1 - 2x = 5
2 x = -4
x = - 2

{x : -2 < x ≤ 0.5 }

Note: The domain of the function refers to the x value while the range of the function refers to the y value.

:thankyou:

i have another question

The diagram shows the graph of the curve y=4/x-1 for x>1

With the aid of the graph http://i47.tinypic.com/2w6w4sk.jpg\"> , find

(a) the range of f(x)=4/x-1 for the domain 2<x<5,
(b) the domain of f defined by f(x)=4/x-1 corresponding to the range 1<f(x)<3.

[Augmum]

LOL123:

Augmum:

[quote=\"LOL123\"]pls help me


The function f is defined by f(x)=1-2x.
(a) Find the range of f corresponding to the domain -1≤x<3
(b) Find the domain of f corresponding to the range 0≤f(x)<5

thanks

Post on behalf on my gal....

(a) subst x = -1 into the equation
y = 1 - 2(-1)
y = 3
Substx = 3 into the equation
y = 1 - 2(3)
y = -5

Thus, the range of f is : { y: -5 < y ≤ 3}

(b) qn means that: find the range of values of x such that f(x) = 0
1 -2x = 0
2x = 1
x = 0.5

1 - 2x = 5
2 x = -4
x = - 2

{x : -2 < x ≤ 0.5 }

Note: The domain of the function refers to the x value while the range of the function refers to the y value.

:thankyou:

i have another question

The diagram shows the graph of the curve y=4/x-1 for x>1

With the aid of the graph http://i47.tinypic.com/2w6w4sk.jpg\"> , find

(a) the range of f(x)=4/x-1 for the domain 2<x<5,
(b) the domain of f defined by f(x)=4/x-1 corresponding to the range 1<f(x)<3.[/quote](a) sub in x= 2
y = 4/2-1
y = 4

Sub in x= 5
y= 4/5-1
y= 1

Range of f(x) : { 1<f(x)<4 }

(b) since f(x) = 4/x-1
And y= f(x),
So, y= 4/x-1
1 < f(x)
1 < 4/x-1
Multiply x-1 on both sides...
x-1 < 4
x < 5

f (x)
So, 4/x-1
Multiply x-1 on both sides...
4 < 3(x-1)
4 < 3x-3
4+3 < 3x
7/3 < x

As x<5 and 7/3<x, the domain is {x: 7/3<x<5}

Hope there is no typo error....

Btw, LOL123, which sec level are u in now ? Is this topic being taught currently ?

[Michaelia0816]

Ty jieheng

[LOL123]

Thank you so much! I have 1 last question


The arrow diagram shows part of the mapping h(x)=12/ax+b, x≠-b/a
Calculate the values of a and b. Hence find

(i) the element that has an image of -2 under this mapping,
(ii) the values of x for which h(x)=x

I am in sec 2 this year. and yup, i am currently doing functions http://i48.tinypic.com/25kq4hs.jpg\">

[shin88]

Pls help me on the following Qin. Thanks

If 600 = 2^3 x 3 x 5^2. Find the smallest value of p such that 600p is a perfect cube.

[jieheng]

shin88:

Pls help me on the following Qin. Thanks

If 600 = 2^3 x 3 x 5^2. Find the smallest value of p such that 600p is a perfect cube.

600 = 2^3 x 3 x 5^2

To be a perfect cube , the index of the prime factors of 600p must be multiple of 3 ,

600p = 2^3*3^3*5^3

p = 3^2*5 = 45

[shin88]

Thanks for your solution. I understand how u get 2^3x3^3x5^3 but how to get 3^2x5 ?

[jieheng]

shin88:

Thanks for your solution. I understand how u get 2^3x3^3x5^3 but how to get 3^2x5 ?

600 = 2^3 x 3 x 5^2

600p = 2^3 * 3^3 * 5^3
2^3 * 3 * 5^2 * p =2^3 * 3^3 * 5^3

p = 3^2*5 = 45

[shin88]

Thanks a lot, I finally understand :rahrah: