help with 'types' of functions...

[zixxer]

Basically,,, the questions go:

1) Identify the type of function: f(x)=5
2) Identify the type of funtion: 2x^3+7x-15


I know this is really simple... but what exactly is he asking for here?

 

[notfred]

1) constant
2) polynomial

 

[zixxer]

ooohhhhhhhhhhh.... okay I was way overanalyzing that


thanks!

 

[zixxer]

okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

 

[Atomicus]

Originally posted by: armatron
okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

1) constant
2) 2nd degree poly
3) same
4) linear (1st degree)

 

[zixxer]

Originally posted by: Atomicus

Originally posted by: armatron
okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

1) constant
2) 2nd degree poly
3) same
4) linear (1st degree)

is 5 constant?

what makes:
f(x)=7x^2-4x+28
second degree, but :
2x^3+7x-15
just a 'polynomial'?

 

[zixxer]

okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

 

[jai6638]

Originally posted by: armatron

Originally posted by: Atomicus

Originally posted by: armatron
okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

1) constant
2) 2nd degree poly
3) same
4) linear (1st degree)

is 5 constant?

what makes:
f(x)=7x^2-4x+28
second degree, but :
2x^3+7x-15
just a 'polynomial'?

Second degree = has an exponent of 2

The 2x^3 + 7x -15 does not have an exponent of 2 and hence is not a second degree..

 

[zixxer]

Originally posted by: jai6638

Originally posted by: armatron

Originally posted by: Atomicus

Originally posted by: armatron
okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

1) constant
2) 2nd degree poly
3) same
4) linear (1st degree)

is 5 constant?

what makes:
f(x)=7x^2-4x+28
second degree, but :
2x^3+7x-15
just a 'polynomial'?

Second degree = has an exponent of 2

The 2x^3 + 7x -15 does not have an exponent of 2 and hence is not a second degree..

so it's, in effect, "third degree"?

Thanks

 

[Atomicus]

Originally posted by: jai6638

Originally posted by: armatron

Originally posted by: Atomicus

Originally posted by: armatron
okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

1) constant
2) 2nd degree poly
3) same
4) linear (1st degree)

is 5 constant?

what makes:
f(x)=7x^2-4x+28
second degree, but :
2x^3+7x-15
just a 'polynomial'?

Second degree = has an exponent of 2

The 2x^3 + 7x -15 does not have an exponent of 2 and hence is not a second degree..

Thanks for pointing it out. I hoped the OP would get the hint of the trend.

BTW, when you compound the interest, maybe you should add that amount you obtained to your initial investment

 

[zixxer]

Originally posted by: Atomicus

Originally posted by: jai6638

Originally posted by: armatron

Originally posted by: Atomicus

Originally posted by: armatron
okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

1) constant
2) 2nd degree poly
3) same
4) linear (1st degree)

is 5 constant?

what makes:
f(x)=7x^2-4x+28
second degree, but :
2x^3+7x-15
just a 'polynomial'?

Second degree = has an exponent of 2

The 2x^3 + 7x -15 does not have an exponent of 2 and hence is not a second degree..

Thanks for pointing it out. I hoped the OP would get the hint of the trend.

BTW, when you compound the interest, maybe you should add that amount you obtained to your initial investment

ahhh... thanks

 

[chuckywang]

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

 

[zixxer]

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

 

[chuckywang]

Originally posted by: armatron
okay this totally isn't in my book.

f(x)=5

f(x)=(x)/(x^2+4)

f(x)=7x^2-4x+28

f(x)=3^(x-1) + 2

f(x)=x+3

1) Constant
2) no real name for this, it's just the quotient of a first degree poly with a second degree poly
3) second degree poly.
4) exponential
5) linear (or first degree poly.)

 

[chuckywang]

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

Yep, subtract off 1500 to get the interest.

 

[zixxer]

Originally posted by: chuckywang

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

Yep, subtract off 1500 to get the interest.

okay... how about the same problem but compounded quarterly rather than continuously?

 

[chuckywang]

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

Yep, subtract off 1500 to get the interest.

okay... how about the same problem but compounded quarterly rather than continuously?

Hmm...been a while, but I think the equation for that is : A = P*(1+r/4)^(4t)

 

[notfred]

What class is this?

 

[z0mb13]

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

I got this answer $1693.359

what I did is 1500*(1+.0625)^2

 

[chuckywang]

Originally posted by: z0mb13

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

I got this answer $1693.359

what I did is 1500*(1+.0625)^2

The problem asks for interest compounded continuously, not compounded annually.

 

[zixxer]

Originally posted by: notfred
What class is this?

calculus. I know all of the 'advanced' stuff... but my teacher didn't actually teach us the first chapter... which is review from basically my soph year in high school...

 

[zixxer]

Originally posted by: chuckywang

Originally posted by: z0mb13

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

I got this answer $1693.359

what I did is 1500*(1+.0625)^2

The problem means compounded continuously, not compounded annually.

yes.. but the next problem is "okay, now compute if interest is compounded quarterly"

and the final might have "yearly, bi-yearly" or who knows

 

[z0mb13]

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: z0mb13

Originally posted by: armatron

Originally posted by: chuckywang

Originally posted by: armatron
okay stuck again on these stupid problems that I learned 4 years ago.. arg

$1500 invested at 6.25% compounded continuously. What is the balance after two years?

A=Pe^(rt)
A=1500e^(.125)
ok


do I then go

A\1500=e^.125
A\1500=ln e^.125
A\1500=.125
A=187.5

Now, I know that $1500 after 2 years at 6.25 doesn't equal... 187.5. so what am I screwing up here?

When you took ln of the e^.125, you didn't do it for the left hand side of the equation.

You pretty much have this thing solved when you wrote A=1500e^(.125). Just do that on your calculator to solve for A.

$1699.72

awesome.. and that includes the initial amount


Thanks man

I got this answer $1693.359

what I did is 1500*(1+.0625)^2

The problem means compounded continuously, not compounded annually.

yes.. but the next problem is "okay, now compute if interest is compounded quarterly"

and the final might have "yearly, bi-yearly" or who knows

then use my formula, divide the percent by four, and use 8 for the period

 

[chuckywang]

Originally posted by: armatron


yes.. but the next problem is "okay, now compute if interest is compounded quarterly"

and the final might have "yearly, bi-yearly" or who knows

see above for my post

 

[Mathlete]

Here is a gereric formula that will work for anything other the continuous(you alreadt have that formula)

A=P * ( 1 + r/n)^nt

where P is the principle
r is the rate given as an APR
n is the number of times interest is compounded per year
t is the number of years

sometimes you will see r/n written as i which stands for the interest per payment period


Also remember that e is a number so when you had 1500e^.0625(2) this is actually a number