Author Topic: FREE KARMA ~ yes another question thread . . ;] (Read 5224 times) Tweet Share

crayolé · « **Reply #30 on:** April 19, 2010, 11:45:05 pm »

Okay, say you're given the function

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

It asks you for the largest domain that makes this a one to one function

would $[-3,\infty)$ be bigger than $(-\infty,-3]$ as it essentially takes 6 extra numbers into account?

Sorry if you dont understand where im coming from :/

m@tty · « **Reply #31 on:** April 20, 2010, 12:09:26 am »

Haha, I've never thought of that before...
I don't think it really matters, either should be accepted.

Anyway, technically both domains are equally infinite: $\infty+6=\infty$ ...

the.watchman · « **Reply #32 on:** April 20, 2010, 05:53:19 am »

We did a question like this at school and I can remember that our teacher preferred us using the one that 'looked bigger' or spanned for more of the plane.

So in this case, I would put down $[-3,\infty]$ , but I don't know if the other answer is wrong.

crayolé · « **Reply #33 on:** April 24, 2010, 06:35:23 pm »

How would you do this?

If $\log_{e} (x) = a$

Then what does $e^{2a}+3e^{a}-2e^{-a}$ equal in terms of x?

Heres what i have so far
$e^{a}=x$
$\therefore$ $e^{2a}=(e^{a})^2=x^2$
$3e^{a}=3x$

I've managed to get this far but i don't know what to do on the last step with the $-2e^{-a}$

Cthulhu · « **Reply #34 on:** April 24, 2010, 06:50:51 pm »

Remember that $e^{\log_e(x)} = x$
then
$e^{2a} = e^{2\log_e(x)} = e^{\log_e(x^2)} = x^2$
blah blah blah for the 2nd term.
$-2e^{-a} = -2e^{-log_e(x)} = -2e^{log_e(x^{-1})} = -\frac{2}{x}$

brightsky · « **Reply #35 on:** April 24, 2010, 07:08:25 pm »

$\because e^a= x$

$\therefore e^{-a} = \frac{1}{e^a} = \frac{1}{x}$ $\text{(Invert both sides)}$

$\text{Hence} -2 e^{-a} = -2 \cdot \frac{1}{x} = -\frac{2}{x}$ $\text{(Multiply both sides by -2)}$

crayolé · « **Reply #36 on:** April 24, 2010, 07:12:41 pm »

Ah i see, thanks guys!

Another question,

Everytime we square root a squared number, do we have to consider whether it was negative or positive?

For example
$\sqrt {\frac {1}{x^2}}$ = $\frac {1}{x}$ or $\frac {-1}{x}$

and when given the formula for a circular function, why do we have to square both sides? Can't we just take care of the square root straight away?
for example
$y=\sqrt {4-x^2}$ to $y={2-x}$

brightsky · « **Reply #37 on:** April 24, 2010, 07:18:21 pm »

Yes for the first one, unless it HAS to be positive (e.g. if you are talking about the length or volume of something, etc).

For the second one, $4 - x^2$ is NOT a perfect square.

$4 - x^2 = (2 - x)(2+x)$

while

$(2 - x)^2 = 4 - 4x + x^2$

crayolé · « **Reply #38 on:** April 24, 2010, 07:40:10 pm »

Ah i see now ;]
Thanks a lot Bright :3

Is there anyway to get from $\frac {x^2.e}{x+2}$ to $\frac {x^2}{e(x+2)}$
because the first one was my answer but the second one was the answer in the book :/

Blakhitman · « **Reply #39 on:** April 24, 2010, 07:43:55 pm »

Quote from: crayola on April 24, 2010, 07:40:10 pm

Ah i see now ;]
Thanks a lot Bright :3

Is there anyway to get from $\frac {x^2.e}{x+2}$ to $\frac {x^2}{e(x+2)}$
because the first one was my answer but the second one was the answer in the book :/

Nope! you're multiplying the expression $\frac {x^2}{x+2}$ by $e$ in the first one and dividing in the second!

the.watchman · « **Reply #40 on:** April 24, 2010, 07:49:40 pm »

Quote from: Blakhitman on April 24, 2010, 07:43:55 pm

Quote from: crayola on April 24, 2010, 07:40:10 pm
Ah i see now ;]
Thanks a lot Bright :3

Is there anyway to get from $\frac {x^2.e}{x+2}$ to $\frac {x^2}{e(x+2)}$
because the first one was my answer but the second one was the answer in the book :/

Nope! you're multiplying the expression $\frac {x^2}{x+2}$ by $e$ in the first one and dividing in the second!

That's right, there must be something wrong in the process

crayolé · « **Reply #41 on:** April 24, 2010, 07:59:15 pm »

The question is Solve for y
$2\log_{e} (x) - \log_{e} (x+2)=1+\log_{e} (y)$

This is how i did it, where did I go wrong?
$\log_{e} (\frac {x^2}{x+2}) =\log_{e} (ey)$

$\frac {x^2}{x+2} = ey$

$\frac {\frac {x^2}{x+2}}{e} = y$

$\frac {x^2}{x+2} \div \frac {e}{1} = y$

$\frac {x^2}{x+2} \times \frac {1}{e} = y$

$\frac {x^2.e}{x+2} = y$

the.watchman · « **Reply #42 on:** April 24, 2010, 08:01:00 pm »

The last two steps:

$\frac{x^2}{x+2} \times \frac{1}{e} = y$

$\frac{x^2}{e(x+2)}= y$

m@tty · « **Reply #43 on:** April 24, 2010, 09:18:12 pm »

Quote from: crayola on April 24, 2010, 07:12:41 pm

Ah i see, thanks guys!

Another question,

Everytime we square root a squared number, do we have to consider whether it was negative or positive?

For example
$\sqrt {\frac {1}{x^2}}$ = $\frac {1}{x}$ or $\frac {-1}{x}$

Unless the square root is both positive AND negative, given by " $\pm$ ", then you only consider the positive square root. So to write that in terms of x you mod it, for your example $\sqrt\left(\frac{1}{x}\right)^2} = \left|\frac{1}{x}\right|=\frac{1}{|x|}=\begin{cases} \frac{1}{x}, x\geq0 \\ -\frac{1}{x},x<0\end{cases}$ .

crayolé · « **Reply #44 on:** May 29, 2010, 04:12:56 pm »

Hey, back again

How do i differentiate these? :/
$f(x)=e^{x}+e^{2}$
$f(x)=4e^{5x}+2x^{2}-e^{-3}$

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Author Topic: FREE KARMA ~ yes another question thread . . ;] (Read 5224 times) Tweet Share

crayolé

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