Pre Calculus help!

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  • #1
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Range of- (x)/((x^2)-9)
please explain in detail

also

the properties of exponents to simplify the expression

((16x^-2)(y^4))^1/2

A bacterial culture starts with 20,000 bacteria, and the number doubles every 40 min. Find the number T of minutes required for the culture to have 50,000 bacteria?
 

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  • #2
Zurtex
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It's been a while since I did domain and range and I'm afraid I might stuff up the explanation but I can help you with some rules of exponents that might help:

[tex](ab)^c = a^cb^c[/tex]

[tex] \left( x^y \right)^z = x^{yz}[/tex]
 
  • #3
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parwana said:
also

the properties of exponents to simplify the expression

((16x^-2)(y^4))^1/2
Here are the properties you want:

[tex](a^x b^y)^z = (a^x)^z (b^y)^z[/tex]

and

[tex](a^x)^y = a^{xy}[/tex]

parwana said:
A bacterial culture starts with 20,000 bacteria, and the number doubles every 40 min. Find the number T of minutes required for the culture to have 50,000 bacteria?
This is going to be an exponential function. Think about how [itex]2^x[/itex] changes when x changes (mentally substitute x=1, x=2, x=3). What if we had a number multiplied by [itex]2^x[/itex]? Think about it, and if you can't figure out where to go, post what you've tried to solve this.
 
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  • #4
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i got the first one and last, but am stuck in the second, this is what i have so far for the second, (1/16x^2)^1/2 X (y^4)^1/2
 
  • #5
arildno
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Zorodious, NO!
The range is not the domain!!

parwana: the range is that set of values the function takes over its domain
 
  • #6
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arildno said:
Zorodious, NO!
The range is not the domain!!
Whoops, that's a pretty big screw-up. Sorry, I zoned out entirely.
 
  • #7
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Re: ((16x^-2)(y^4))^1/2

parwana, was the original problem this:

[tex][(16x)^{-2}y^4]^{1/2}[/tex]

or was it this?

[tex][16x^{-2}y^4]^{1/2}[/tex]

In the former case, the 16 is also raised to the -2 power, in the latter case, only the x is.

Here's one way to solve the first one:

[tex][(16x)^{-2}y^4]^{1/2} = {((16x)^{-2})}^{1/2} {(y^4)}^{1/2} = (16x)^{-2 \cdot 1/2} y^{4 \cdot 1/2} = (16x)^{-1} y^2 = \frac {y^2}{16x}[/tex]

Here's another way:

[tex][(16x)^{-2}y^4]^{1/2} = \sqrt {(16x)^{-2}y^4} = \sqrt {\frac{y^4}{(16x)^2}} = \frac {\sqrt {y^4}}{\sqrt {(16x)^2}} = \frac {y^2}{16x}[/tex]

For the second possibility, here is one potential solution:

[tex][16x^{-2}y^4]^{1/2} = \sqrt {16 x^{-2} y^4} = 4 x^{-1} y^2 = \frac {4 y^2}{x}[/tex]
 
  • #8
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Thanks everyone, it was the latter one zorodius. Here are some more questions!

Given that f(x)= 10, find the ratio (f(x+t)-f(x))/t

Find the roots of 4/h = (3/h^2)+1

One pipe can fill a swimming pool in 5 hours less than another. Together they fill the swimming pool in 5 hours. How long would it take each pipe to fill the tank alone?

Given that f(x)= cos(x+1) and g(x) = ((x^2)-1)/2, find (f o g)(x)
 
  • #9
arildno
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Post your thoughts on how to attack the first exercise (or if you don't know how, specify what's troubling you)
 
  • #10
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for the first one i think it would be ((10+t)-(10)/t), now what??
 
  • #11
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see for urself its 1

,post ur try for all the question
 
  • #12
arildno
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parwana:
That is incorrect.

You have been given that f(x)=10
This means, that whatever number you plug into f, the value you get out is 10
(Ok?)
This is an example of a constant function.

Now, let us consider what is meant by the expression f(x+t):

If x and t are numbers, then surely, x+t is a number as well!

But we know, that whatever number we plug into f will give the output 10
Since x+t is some number, we have f(x+t)=10

Please respond and say what,if anything, you don't understand about this answer.
 
  • #13
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I dont get it, i mean i can see where the answer is 1, cause the 10's cancel out and t/t=1
 
  • #14
Gokul43201
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parwana, this thread belongs in the homework help section. And you will receive help on a problem only if you show what you've tried and where you're stuck.

The idea of this forum is to clarify your doubts, not to do your work for you.

That said - and keeping arildno's explanation in mind - understand where you went wrong in your attempt. There is no rule that says : f(x+t) = f(x) + f(t) OR, even worse, f(x+t) = f(x) + t. The latter is what you applied in your attempt.

Show us what you've done on the other problems too...and we'll help you with them.

himanshu, how do you say it's 1 ? I think not. The derivative (slope) of a constant function ?
 
  • #15
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parwana said:
Thanks everyone, it was the latter one zorodius. Here are some more questions!

Given that f(x)= 10, find the ratio (f(x+t)-f(x))/t
What you want to do here is to figure out what you can write instead of f(x+t) and f(x). For instance, you were given that f(x) = 10, so you can write the number 10 instead of f(x) in the problem. You can also figure out that f(x+t) is 10 as well, because f(x) doesn't change no matter what you put between the parentheses!

parwana said:
Find the roots of 4/h = (3/h^2)+1
I'll start this off:

[tex]\frac {4}{h} = \frac {3}{h^2} + 1[/tex]

[tex]0 = \frac {3}{h^2} + 1 - \frac {4}{h}[/tex]

[tex]0 = 3 + h^2 - 4h = h^2 - 4h + 3[/tex]

Can you see where to go from there?

parwana said:
One pipe can fill a swimming pool in 5 hours less than another. Together they fill the swimming pool in 5 hours. How long would it take each pipe to fill the tank alone?
First, let's give names to the quantities we're talking about.

A is the rate at which the fast pipe fills up pools. (in pools / hour)
B is the rate at which the slow pipe fills up pools. (in pools / hour)
Ta is the time it would take A to fill up a whole pool. (in hours)
Tb is the time it would take B to fill up a whole pool. (in hours)

Then "the rate at which a pipe fills up pools" times "the number of hours it takes that pipe to fill up a pool" should be equal to one pool, and so:

[tex]1 = A \cdot T_a[/tex]
[tex]1 = B \cdot T_b[/tex]

We also know, from the problem, that the time for the fast pipe to fill up a pool is five hours less than the time for the slow pipe to fill up a pool, so:

[tex]T_a = T_b - 5[/tex]

And we know that, working together, they fill one pool in five hours. Working together, their total rate is the sum of their individual rates, and:

[tex]1 = (A + B) 5[/tex]

So A + B is the rate of both pipes working together, and 5 is the number of hours it takes them to fill one pool.

Can you see where to go from there?

parwana said:
Given that f(x)= cos(x+1) and g(x) = ((x^2)-1)/2, find (f o g)(x)
I'll show an example of how a similar problem could be solved:

Given that a(x) = 2x and b(x) = x + 1, find (a o b)(x).

We replace (a o b)(x) with the equivalent a(b(x)). For the innermost function, we know that b(x) = x + 1, so a(b(x)) = a(x+1).

a(x) = 2x, meaning that to evaluate function a, we multiply whatever is between the parentheses by two. We had a(x+1), so a(x+1) = 2 * (x+1) = 2x + 2.

Your problem can be solved in the same way that this example was.
 
  • #16
arildno
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parwana: I think your basic misunderstanding lies in what we mean by a function:

1. f(x)=10
When you're given this, it means that for every number you chose to substitute for x, you'' get 10 back.
For example, f(1)=10, f(3.14)=10

When you get an equation like this, x is often called "the variable"

Now, I would like to ask you:
If I put an arbitrary number "t" into f, what do I get then (What is the value of f(t))?
 
  • #17
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arildno f(t)= 10, but then Given that f(x)= 10, find the ratio (f(x+t)-f(x))/t

so then it will be (10-10)/t??

Thanks for the help zorodius, i understand the latter 2, but this oen still troubles me
 
  • #18
arildno
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Yes that is correct.

Be particularly aware of that when f(x+t) and f(x) are used in the ratio, x,x+t is meant to be arbitrary numbers, and not as a "variable" used in the definition of f.
So, (10-10)/t=0, for all (non-zero) t, since 10-10=0
 
  • #19
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one last question, this has been troubling my mind

Given that f(s)= 4s^2+3s+8, find the ratio (f(t+h)-f(t))/h


it gives s, but in the ratio it uses a t and h?
 
  • #20
arildno
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Treat t+h and t as arbitrary numbers to be plugged in where s stands in the definition of f.
For example, we must have:
f(t+h)=4(t+h)^2+3(t+h)+8
 
  • #21
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so will it be

((4s^2+3s+8)-(4s^2+3s+8))/h????

but then it will be zero??
 
  • #22
arildno
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No!
Even though t+h and t is arbitrary numbers, they are only equal if h is 0.
We have:
f(t)=4t^2+3t+8
 
  • #23
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parwana said:
one last question, this has been troubling my mind

Given that f(s)= 4s^2+3s+8, find the ratio (f(t+h)-f(t))/h


it gives s, but in the ratio it uses a t and h?
The letter s is anything you want it to be. Think of it like saying, "Fill in the blank: f(___) = 4(___)^2 + 3(___) + 8." So if you wanted to put in the number 2, for example, you'd have f(2) = 4(2)^2 + 3(2) + 8 = 30. The value of the function f(s) at s=2 is 30. If you wanted to put in the number "t plus h", you'd have:

[tex]f(t+h) = 4(t+h)^2 + 3(t+h) + 8[/tex]

[tex]= 4(t^2 + 2th + h^2) + 3t + 3h + 8[/tex]

[tex]= 4t^2 + 8th + 4h^2 + 3t + 3h +8[/tex]
 
  • #24
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so will this be it

((4(t+h)^2 + 3(t+h)+8)-((4t^2)+3t+8))/ h?????

is that right, and also why cant u do f(h)= 4h^2......
 
  • #25
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parwana said:
so will this be it

((4(t+h)^2 + 3(t+h)+8)-((4t^2)+3t+8))/ h?????
Yes, that's right. You can and probably should simplify that expression in your answer.

parwana said:
and also why cant u do f(h)= 4h^2......
I don't understand your question - you could put an h into f(s) if you wanted to, there's nothing wrong with that. It just wouldn't help you solve the problem.
 

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