Thread: Need help with ff.

Hello all!

I'm not really good in math... recently, our prof announced we would be having a test in the coming week, and he gave us the ff. problems as review questions... trouble is, I'm having a hell of a time with it! Please help!

1) f(g(x)) = x+5/x-7 and g(x)=2x+1/3; what is f(x), and find f(x+1/X-1)

2) A circle is inside a rectangle, and it is tangent to each of the longer sides at the Midpoint. Area of rectangle - Area of circle = 126 sq. cm, and the sum of their perimeters is 112cm. What are the dimensiosn of the rectangle? Use pi=22/7.

3) Find x and y so that: their sum x sum of their squares is 5500; their difference x difference of their squares is 352.

4) A cable from a suspension bridge hands in the form of a parabola, thereby distributing the load horizontally. The distance between one tower and another is 150m; the cables are 22m above the roadway (highest pt.); and te lowest pt. above the roadway is 7m.

What is the vertical distance to the cable from a point on the roadway, 15m from the tower?

Thanks everyone!!!

Originally Posted by umbraculum

...
2) A circle is inside a rectangle, and it is tangent to each of the longer sides at the Midpoint. Area of rectangle - Area of circle = 126 sq. cm, and the sum of their perimeters is 112cm. What are the dimensiosn of the rectangle? Use pi=22/7....

Hello,

let l be the length of the rectangle
and r be the radius of the circle then you get the equations according to your problem:





(Unfortunately the Latex isn't working anymore. So I repeat my equations in plain text)

equ(1): A_{rectangle}-A_{circle}: l * 2r - π* r^2=126

equ(2): p_{rectangle}+p_{circle}: 2l+4r+2πr=112

Now calculate l from equ(1) using π = 22/7 : l = (126+22/7*r^2)/(2r)
and put this term into equ(2) instead of l:

2*(126+22/7*r^2)/(2r) +4r+2πr=112

After a few transformation you'll get a quadratic equation:

94r^2 - 784r + 882 = 0

which has 2 solutions: r = 7 or r = 63/47

Now plug in these values to calculate l. You get:
l = 20 or l = 2308/47

EB

Originally Posted by umbraculum

...

4) A cable from a suspension bridge hands in the form of a parabola, thereby distributing the load horizontally. The distance between one tower and another is 150m; the cables are 22m above the roadway (highest pt.); and te lowest pt. above the roadway is 7m.
What is the vertical distance to the cable from a point on the roadway, 15m from the tower?...

Hello,

use a coordinate system. Place the origin in the middle between the two towers at road level. Then the parabola has the general equation:

p(x) = a*x^2 + 7

You know that the height of the cable in a distance of 75 m from the origin is 22 m. Plug in these values to calculate a:

22 = a * (75)^2 + 7 <===> a = 15/5625 = 1/375

The complete equation of the parabola is:

p(x) = 1/375*x^2 + 7

15 m from the tower means that the x-value is 60. Plug in this value into the equation of the parabola:

p(60) = 1/375*(60)^2+7 = 16.6 m

I've attached a diagram which shows the situation.

EB

Originally Posted by umbraculum

...Please help!

1) f(g(x)) = x+5/x-7 and g(x)=2x+1/3; what is f(x), and find f(x+1/X-1)
...

Hello,

if f(x) is the term of a function then f^(-1)(x) is the term of the inverse function if it actually exists.
You ought to know that f(f^(-1)(x)) = f^(-1(f(x)) = x

1. step: Calculate g^(-1)(x) = 1/2*x-1/6

2. f(g(g^(-1)(x))) = f(x)

Therefore: f(x) = (1/2*x - 1/6 + 5)/(1/2*x - 1/6 - 7) = (1/2*x + 29/6)/(1/2*x - 43/6) = (x + 29/3)/(x - 43/3)

Now that you know f(x) you can calculate f((x+1)/(x-1)):

f((x+1)/(x-1)) = ((x+1)/(x-1) + 29/3)/((x+1)/(x-1) - 43/3) =
(x+1 + 29/3*x - 29/3)/(x-1))/(x+1 - 43/3*x + 43/3)/(x-1)) =
(32/3*x - 26/3)/(-40/3*x + 46/3) = (32x-26)/(-40x+46) = (16x-13)/(-20x+26)

EB

Hello,umbraculum!

This will be hard to explain without LaTeX . . .

1) f(g(x)) = (x + 5)/(x - 7) and g(x) = (2x + 1)/3.

(a) Find f(x)

(b) Find f([x+1]/[x-1])

Assume that: . f(x) .= .(ax + b)/(cx + d)

. . . . . . . . . . . . . . .a[(2x+1)/3] + b
Then: . f(g(x)) . = . --------------------
. . . . . . . . . . . . . . .c[(2x+1)/3] + d

. . . . . . . . . . . . . . . . . . . . . . .2ax + a + 3b
Multiply top and bottom by 3: . -----------------
. . . . . . . . . . . . . . . . . . . . . . .2cx + c + 3d


This is equal to (x + 5)/(x - 7)

. . . . . . . . . . 2ax + (a + 3b) . . . x + 5
so we have: . ------------------ .= .-------
. . . . . . . . . . 2cx + (c + 3d) . . . .x - 7


Equate coefficients: . 2a .= .1 . . a + 3b .= .5
. . . . . . . . . . . . . - - 2c .= .1 . . c + 3d .= .-7

. . and we have: . a = 1/2, . b = 3/2
. . . . . . . . . . - - -c = 1/2, . c = -5/2


. . . . . . . . . . . . (1/2)x + 3/2 . . . . x + 3
Then: . f(x) . = . --------------- . = . -------
. . . . . . . . . . . . (1/2)x - 5/2 . . . . .x - 5

Hello again, umbraculum!

I found a solution to #3 . . .
. . but there must be a better way!

3) Find x and y so that:
their sum times the sum of their squares is 5500.
their difference times the difference of their squares is 352.

We have: . (x + y)(x² + y²) .= .5500 . . → . . x³ + x²y + xy² + y³ .= .5500 . [1]
. . . .And: . .(x - y)(x² - y²) . = . 352 . . .→ . . .x³ - x²y - xy² + y³ . = . 352 . [2]

. . . Add [1] and [2]: . . 2x³ + 2y³ . = .5852 . . → . . .x³ + y³ . = . 2926 . [3]
Subtract [1] and [2]: . 2x²y + 2xy² .= .5148 . . → . . x²y + xy² .= .2574 . [4]

We have: . (x + y)(x² - xy + y²) .= .2926 . [5]
. . . .and: . . . . - - xy(x + y) . . . = .2574 . [6]

. . . . . . . . . . . . . .(x + y)(x² - xy + y²) . . . .2926 . . . . . . x² - xy + y² . . .113
Divide [5] by [6]: . ------------------------ . = . ------ . . → . . -------------- .= .-----
. . . . . . . . . . . . . . . . .xy(x + y) . . . . . . . . .2574 . . . . . . . . x + y . . . . . 117

This simplifies to the quadratic: .117x² - 250xy + 117y² .= .0

. . which factors: .(13x - 9y)(9x - 13y) .= .0

. . and has roots: . y .= .9x/13 and 13x/9


Substitute y = 9x/13 into [3]: .x³ + (9x/13)³ .= .2926

. . which gives us: .x = 13, y = 9


Substitute y = 13x/9 and we get: .x = 9, y = 13

Well, my solution to 3 isn't any shorter than Soroban's, but I feel it's a bit more straightforward.

It starts the same way as Soroban's:
(x + y)(x^2 + y^2) = 5500
(x - y)(x^2 - y^2) = 352

Expanding both we get:
x^3 + x^2y + xy^2 + y^3 = 5500
x^3 - x^2y - xy^2 + y^3 = 352

Add both equations:
2x^3 + 2y^3 = 5852 ==> x^3 + y^3 = 2926

Subtract both equations:
2x^2y + 2xy^2 = 5148 ==> x^2y + xy^2 = 2574

Now we take a turn away from Soroban's solution.

Note that
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = (x^3 + y^3) + 3(x^2y + xy^2)

So
(x + y)^3 = 2926 + 3*2574 = 10648 ==> x + y = 22

Thus
x + y = 22 and y = 22 - x

Now take a new look at your first condition:
(x + y)(x^2 + y^2) = 5500

(22)(x^2 + [22 - x]^2) = 5500

(22)(2x^2 - 44x + 484) = 5500

2x^2 - 44x + 484 = 250

x^2 - 22x + 242 = 125

x^2 - 22x + 117 = 0

(x - 9)(x - 13) = 0

Thus x = 9, 13 which gives y = 13, 9 respectively.

-Dan

Lovely work, Dan!

I tried to find ways to square/cube various expressions
. . as you did, but didn't get very far.

Nice going!

Originally Posted by Soroban

Lovely work, Dan!

I tried to find ways to square/cube various expressions
. . as you did, but didn't get very far.

Nice going!

I'm blushing!

-Dan