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3 years ago

Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.

Mathematics

1 answer:

Eduardwww [97]3 years ago

8 0

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

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Determine the truth value of each of these statements if thedomainofeachvariableconsistsofallrealnumbers.

hoa [83]

Answer:

a)TRUE

b)FALSE

c)TRUE

d)FALSE

e)TRUE

f)TRUE

g)TRUE

h)FALSE

i)FALSE

j)TRUE

Step-by-step explanation:

a) For every x there is y such that $x^2=y$ :

TRUE

This statement is true, because for every real number there is a square number of that number, and that square number is also a real number. For example, if we take 6.5, there is a square of that number and it equals 39.0625.

b) For every x there is y such that $x=y^2$ :

FALSE

For example, if x = -1, there is no such real number so that its square equals -1.

c) There is x for every y such that xy = 0

TRUE

If we put x = 0, then for every y it will be xy=0*y=0

d)There are x and y such that $x+y\neq y+x$

FALSE

There are no such numbers. If we rewrite the equation we obtain an incorrect statement:

$x+y \neq y+x\\x+y - y-y\neq 0\\0\neq 0$

e)For every x, if $x \neq 0$ there is y such that xy=1:

TRUE

The statement is true. If we have a number x, then multiplying x with 1/x (Since x is not equal to 0 we can do this for ever real number) gives 1 as a result.

f)There is x for every y such that if $y\neq 0$ then xy=1.

TRUE

The statement is equivalent to the statement in e)

g)For every x there is y such that x+y = 1

TRUE

The statement says that for every real number x there is a real number y such that x+y = 1, i.e. y = 1-x

So, the statement says that for every real umber there is a real number that is equal to 1-that number

h) There are x and y such that

$x+2y = 2\\2x+4y = 5$

We have to solve this system of equations.

From the first equation it yields x=2-2y and inserting that into the second equation we have:

$2(2-2y)+4y=5\\4-4y+4y=5\\4=5$

Which is obviously false statement, so there are no such x and y that satisfy the equations.

FALSE

i)For every x there is y such that

$x+y=2\\2x-y=1$

We have to solve this system of equations.

From the first equation it yields $x=2-y$ and inserting that into the second equation we obtain:

$2(2-y)-y=1\\4-2y-y=1\\4-3y=1\\-3y=1-4\\-3y=-3\\y=1$

Inserting that back to the first equation we obtain

$x=2-1\\x=1$

So, there is an unique solution to this equations:

x=1 and y=1

The statement is FALSE, because only for x=1 (and not for every x) exists y (y=1) such that

$x+y=2\\2x-y=1$

j)For every x and y there is a z such that

$z=\frac{x+y}{2}$

TRUE

The statament is true for all real numbers, we can always find such z. z is a number that is halway from x and from y.

5 0

3 years ago

PLEASE HELP

Licemer1 [7]

Answer:

35 dimes and 60 nickels

Step-by-step explanation:

let d = # of dimes and n = # of nickels

We know that Timmy has a total of 95 coins, consisting of dimes and nickels

So d + n = 95

The total value of his bank is $6.50

A dime is worth $0.10 and a nickel is work $0.05

So 0.10d + 0.05n = $6.50

Note that we've just created a system of equations that we can solve

We have the two equations

0.10d + 0.05n = 6.50 and d + n = 95

There are many different methods we can use to solve this system but the easiest way in this situation is probably going to be the substitution method.

First we are going to want to rearrange the terms of the second equations so that one of the variables are defined.

d + n = 95

- subtract d from both sides -

n = 95 - d

We now defined "n"

Now that we have defined one of the variables we can plug in ( or substitute ) the value of it into the other equation. Once we substitute it we can solve for the other variable.

0.10d + 0.05n = 6.50

n = 95 - d

0.10d + 0.05(95 - d) = 6.50

we now solve for d

0.10d + 0.05(95 - d) = 6.50

step 1 distribute the 0.05

0.05 * 95 = 4.75 and 0.05 * -d = -0.05d

0.10d + 4.75 - 0.05d = 6.50

step 2 combine like terms 0.10d - 0.05d = 0.05d

0.05d + 4.75 = 6.50

step 3 subtract 4.75 from both sides

0.05d = 1.75

step 4 divide both sides by 0.05

0.05d / 0.05 = d and 1.75 / 0.05 = 35

d = 35

Now that we have found the value of one variable we can plug it into one of the equations and solve for the other variable ( note that the equation that we use does not matter, we will acquire the same answer )

d + n = 95

d = 35

35 + n = 95

subtract 35 from both sides

n = 60

We can conclude that he has 35 dimes and 60 nickels

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snow_tiger [21]

Answer:

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Step-by-step explanation:

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The 17 students on the math team want to raise $483.62 to buy practice books they have already raised $218.25 of each students r

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Answer:

Each stuydent must raise $15.61 to reach their goal.

Step-by-step explanation:

Their goal is 483.62, but they already have 218.25 so we subtract,

483.62 - 218.25 = 265.37

Since there are 17 students and they raise the same amount we need to divide by the amount of students to find out how much money each of them need to raise to add up to their goal.

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