We love math

[arzafar]

what is the limit of y when y-> -3+ (y^5 - y^3)/((y^2) -9)?
when y tends to -3 by the right....

the question is not clear, can you please rewrite it.

 

[Abdul25]

what is the limit of y when y-> -3+ (y^5 - y^3)/((y^2) -9)?
when y tends to -3 by the right....

answer is infinity .. , as far i could try..

 

[jahid89]

answer is infinity .. , as far i could try..

hummm ur answer is correct but not complete...

 

[jahid89]

what is the limit of y when y-> -3+ (y^5 - y^3)/((y^2) -9)?
when y tends to -3 by the right....

Lim x→-3+ [ (x^5 - x^3) / (x^2 -9) ] = Lim x→-3+ [ f(x)/g(x) ]

L'Hospital is not needed since f(-3) = ∞, g(-3) = 0
So we directly conclude
Lim x→-3+ [ f(x)/g(x) ] = ∞ =undefined

 

[jahid89]

next question.....

what is the domain of f(z)=3/4-z^2+log(base)10(z^3-z)?

i.e
f(z)=(3/4)-z^2+log(base)10(z^(3)-z)

 

[arzafar]

next question.....

what is the domain of f(z)=3/4-z^2+log(base)10(z^3-z)?

o yaar halka hoja.
i mean dont use so many symbols because we cant use mathematical notation and this can lead to confusion. your previous question also wasnt entirely clear.

for eg. z^3-z could mean z raised to the power 3 and then minus z
or it could also mean z raised to the power 3-z.

 

[jahid89]

o yaar halka hoja.
i mean dont use so many symbols because we cant use mathematical notation and this can lead to confusion. your previous question also wasnt entirely clear.

for eg. z^3-z could mean z raised to the power 3 and then minus z
or it could also mean z raised to the power 3-z.

oops ...cool down.... samjh gya yaar
my question is..

find domain... f(z)=(3/4)-z^2+log(base)10(z^(3)-z)

 

[arzafar]

oops ...cool down.... samjh gya yaar
my question is..

find domain... f(z)=(3/4)-z^2+log(base)10(z^(3)-z)

lemme give this one a try

f(z) = X + Y
where X = (3/4)-z^2 and Y = lg[z^3 - z]

domian (X) = {a: a is real}

Y = lg[z(z^2 - 1)] = lg [z(z-1)(z+1)]
domain (Y) = {a: a>0 and a is not= 1}

domian f(z) = domain (X) n domain (Y) {n - intersection}
domain f(z) = {a: a>0 and a is not= 1}

 

[jahid89]

lemme give this one a try

f(z) = X + Y
where X = (3/4)-z^2 and Y = lg[z^3 - z]

domian (X) = {a: a is real}

Y = lg[z(z^2 - 1)] = lg [z(z-1)(z+1)]
domain (Y) = {a: a>0 and a is not= 1}

domian f(z) = domain (X) n domain (Y) {n - intersection}
domain f(z) = {a: a>0 and a is not= 1}

hummmmmm ... m not satisfy with ur answer
domain is not define properly...

 

[jahid89]

next question.....

what is the domain of f(z)=3/4-z^2+log(base)10(z^3-z)?

i.e
f(z)=(3/4)-z^2+log(base)10(z^(3)-z)

consider the equetion....
when z^3 - z < = 0 the f( z ) would be meaning less and undefined log(0)

z^3 - z ! = 0 and

=> z ( z^2 - 1 ) ! = 0

=> z ( z + 1 ) ( z - 1 ) ! = 0

=> z ! = 0 and z + 1 ! = 0 and z -1 ! = 0

=> z ! = 0 and z ! = -1 and z ! = 1

and z^3 - z > 0

=> z ( z^2 - 1 ) > 0

=> z > 0 and z > 1 and z > -1 = > z should be greater than -1

so domain of f ( z ) is ( -1 , 0 ) u ( 1 , 2 ] u [ 2 , ∞ )

 

[arzafar]

consider the equetion....
when z^3 - z < = 0 the f( z ) would be meaning less and undefined log(0)

z^3 - z ! = 0 and

=> z ( z^2 - 1 ) ! = 0

=> z ( z + 1 ) ( z - 1 ) ! = 0

=> z ! = 0 and z + 1 ! = 0 and z -1 ! = 0

=> z ! = 0 and z ! = -1 and z ! = 1

and z^3 - z > 0

=> z ( z^2 - 1 ) > 0

=> z > 0 and z > 1 and z > -1 = > z should be greater than -1

so domain of f ( z ) is ( -1 , 0 ) u ( 1 , 2 ] u [ 2 , ∞ )

oh man well i basically messed up the final answer

 

[jahid89]

next proble,....
Now comes the difficult part
y=? does the graph of f (y) have a horizontal tangent? f (y) = 3 (y^3) + 7 (y^2) + 3 (y) + 6

 

[arzafar]

next proble,....
Now comes the difficult part
y=? does the graph of f (y) have a horizontal tangent? f (y) = 3 (y^3) + 7 (y^2) + 3 (y) + 6

is "y=?" a question? anyways

differentiate f(y) wrt y
f'(y) = 9y^2 +14y + 3

at the horizontal tangent f '(y) = 0
=> 9y^2 + 14y + 3 = 0
=> y^2 + (14/9)y + 1/3 = 0
=> (y + 7/9)^2 - (7/9)^2 + 1/3 = 0
=> (y+ 7/9)^2 = 49/81 - 27/81
=> (y+ 7/9)^2 = 22/81
=> y = -7/9 +/-(22/81)^(1/2)

 

[jahid89]

is "y=?" a question? anyways

differentiate f(y) wrt y
f'(y) = 9y^2 +14y + 3

at the horizontal tangent f '(y) = 0
=> 9y^2 + 14y + 3 = 0
=> y^2 + (14/9)y + 1/3 = 0
=> (y + 7/9)^2 - (7/9)^2 + 1/3 = 0
=> (y+ 7/9)^2 = 49/81 - 27/81
=> (y+ 7/9)^2 = 22/81
=> y = -7/9 +/-(22/81)^(1/2)

y = -7/9 +/-(22/81)^(1/2)= -1.298935084
correct..
good brother:jumpclap:

 

[arzafar]

y = -7/9 +/-(22/81)^(1/2)= -1.298935084
correct..
good brother:jumpclap:

phew i was worried for a minute

 

[arzafar]

you have 64 blue chips. While waiting in line at the casino to cash them in you decide to bet half that number on the toss of a coin. Win or lose, you bet half of your new total on a second toss; then half of the new total on a third toss; and so on, until you reach the cashier's window. Say you make 6 tosses in all, winning three and losing three. Do you come out even, ahead or behind and in case you come out ahead or behind by what amount?

 

[jahid89]

i think behind 32

 

[arzafar]

you lose 37 chips. It doesn't matter in what order you win or lose as long as you win 3 and lose 3 bets.

for eg.

1) bet 32; lose 32 ; total = 32 + 0 = 32
2) bet 16; win 32 ; total = 16 + 32 = 48
3) bet 24; lose 24; total = 24 + 0 = 24
4) bet 12; win 24; total = 12 + 24 = 36
5) bet 18; win 36; total = 18 + 36 = 54
6) bet 27; lose 27; total = 27 + 0 = 27

money lost = 64 - 27 = 37

 

[jahid89]

you lose 37 chips. It doesn't matter in what order you win or lose as long as you win 3 and lose 3 bets.

for eg.

1) bet 32; lose 32 ; total = 32 + 0 = 32
2) bet 16; win 32 ; total = 16 + 32 = 48
3) bet 24; lose 24; total = 24 + 0 = 24
4) bet 12; win 24; total = 12 + 24 = 36
5) bet 18; win 36; total = 18 + 36 = 54
6) bet 27; lose 27; total = 27 + 0 = 27

money lost = 64 - 27 = 37

oops :shymuslima1: littile bit mistake

 

[arzafar]

oops :shymuslima1: littile bit mistake

you're a bro right?