Author Topic: Dekoyl's question thread (Read 22040 times) Tweet Share

ell · « **Reply #105 on:** November 05, 2008, 11:04:08 pm »

Quote from: dekoyl on November 05, 2008, 10:56:26 pm

1. I think I asked in someone's thread but I forgot which one. I wanted to know if adding "+c" when asked to find an antiderivative would technically be wrong?

2.What is the range of $1 - 9sin^{2}(2x)$ with a domain of $0\leq x \leq \pi$ ? I'm not too sure how to work it out as I'm unfamiliar with the graph's shape. This a non-calculator question.

Thanks.

1. If it asks for an antiderivative, you should state that c=0, or at least say 'an antiderivative of [insert function] is [insert antiderivative without c]'

2. The range of $\sin(2x)$ over $0\leq x \leq \pi$ is $[-1,1]$ , so the range of $\sin^{2}(2x)$ is $[0,1]$ i.e. if you think about it like this:

$\left(\sin(2x)\right)^{2}$

since $\sin(2x)$ can only take values between -1 and 1, the range of the square of it can only produce values between 0 and 1.

fredrick · « **Reply #106 on:** November 05, 2008, 11:05:48 pm »

Yeh just put c=0 just incase

dekoyl · « **Reply #107 on:** November 05, 2008, 11:38:00 pm »

Thanks ell.

dekoyl · « **Reply #108 on:** November 06, 2008, 11:31:12 am »

If a question says:
"State the rule for $f^{-1}$ ...."
Do you give the answer as:
$f^{-1}(x) =$ ...without the domain?

Thanks.

Mao · « **Reply #109 on:** November 06, 2008, 11:42:26 am »

always give domain of inverse functions

dekoyl · « **Reply #110 on:** November 06, 2008, 12:57:18 pm »

Quote from: Mao on November 06, 2008, 11:42:26 am

always give domain of inverse functions

Yep but if it asks for the rule, do you give the domain as well? Our teacher said only to give the domain if it asks for a function.

/0 · « **Reply #111 on:** November 06, 2008, 01:07:08 pm »

Quote from: dekoyl on November 06, 2008, 12:57:18 pm

Quote from: Mao on November 06, 2008, 11:42:26 am
always give domain of inverse functions
Yep but if it asks for the rule, do you give the domain as well? Our teacher said only to give the domain if it asks for a function.

No

"When the inverse function is asked for, the domain must be given. Students will be penalised in the future if the domain is left out. If only the rule for the inverse function is asked for, the domain does not have to be given."

dekoyl · « **Reply #112 on:** November 06, 2008, 01:27:48 pm »

Thanks for that /0. That clears things up.

dekoyl · « **Reply #113 on:** November 06, 2008, 03:12:03 pm »

This was on bec's thread but I didn't understand Mao's answer

Things don't click as easily for me. (And where's this from? )

Let $a = e^{x}$

$a^{2} + ba + 1 = 0$

No real solutions $\implies \bigtriangleup = b^{2} - 4 < 0$

$b < 2$ or $b > -2$

I also know that $a > 0$ as $e^{x} > 0$

$a = \frac{-b \pm \sqrt{b^{2} - 4}}{2} > 0$

Other than that I'm not sure how to arrive at answer B or how to include 2 as a solution.

Thanks.

dekoyl · « **Reply #114 on:** November 06, 2008, 06:14:19 pm »

Shameless bump.

And also:

2. If say, $\frac{1}{x-1}$ was restricted to $x\in (3,9)$ , would the asymptote of $x = 1$ still need to be drawn when sketching it?

/0 · « **Reply #115 on:** November 06, 2008, 06:30:10 pm »

We need only be bothered with the case $-b+\sqrt{b^2-4} \leq 0$ . If we complete, this, the other case will follow immediately.

STEP 1

$-b+\sqrt{b^2-4} \leq 0$

$\Rightarrow \sqrt{b^2-4} \leq b$

But since $\sqrt{b^2} = b$ if $b$ is positive, it follows that $\sqrt{b^2-4} \leq b$ for all positive values of $b$ .

(Notice that the square root is only defined for $b\leq -2, b \geq 2$ , so in step 1 we have found that $b\in [2, \infty)$ )

STEP 2

OR we require $b^2-4 < 0 \Rightarrow -2<b<2$

Hence the required set of values is $(-2, \infty)$

Mao · « **Reply #116 on:** November 06, 2008, 06:31:49 pm »

first question:

I advice graphing $y=\frac{-x+\sqrt{x^2-4}}{2}$ and $y=\frac{-x-\sqrt{x^2-4}}{2}$ , then find where y is positive in both. the union of the set is what "b" can be.

2. no.

dekoyl · « **Reply #117 on:** November 06, 2008, 06:44:41 pm »

Quote from: DivideBy0 on November 06, 2008, 06:30:10 pm

We need only be bothered with the case $-b+\sqrt{b^2-4} \leq 0$ . If we complete, this, the other case will follow immediately.

Sorry /0 but why is only $-b+\sqrt{b^2-4} \leq 0$ considered? I'm not too sure why there's $\leq 0$ as I thought $\frac{-b \pm \sqrt{b^{2} - 4}}{2} > 0$ because $e^{x} > 0$

And thanks Mao. I'm still stuck algebraically.

/0 · « **Reply #118 on:** November 06, 2008, 06:50:05 pm »

Quote from: dekoyl on November 06, 2008, 06:44:41 pm

Quote from: DivideBy0 on November 06, 2008, 06:30:10 pm
We need only be bothered with the case $-b+\sqrt{b^2-4} \leq 0$ . If we complete, this, the other case will follow immediately.
Sorry /0 but why is only $-b+\sqrt{b^2-4} \leq 0$ considered? I'm not too sure why there's $\leq 0$ as I thought $\frac{-b \pm \sqrt{b^{2} - 4}}{2} > 0$ because $e^{x} > 0$

And thanks Mao. I'm still stuck algebraically.

We need for there to be NO real solutions.

i.e. $e^x \leq 0 \Rightarrow \frac{-b \pm \sqrt{b^2-4}}{2} \leq 0$ .

Also, every time $-b +\sqrt{b^2-4} \leq 0$ , then it must be the case that $-b-\sqrt{b^2-4} \leq 0$ , since $-b-\sqrt{b^2-4} \leq -b+\sqrt{b^2-4}$ (because of the negative sign). We need them to both be 0 or negative, so by ensuring the + case we get them both at the same time.

Mao · « **Reply #119 on:** November 06, 2008, 06:52:13 pm »

case 1, $b\ge 0$

$\implies -b \le 0$

since A need to be greater than 0, we hence can only use the positive strand.

$A=\frac{-b+\sqrt{b^2-4}}{2}$

$\sqrt{b^2-4}\le \sqrt{b^2} = |b|$

$A = \frac{-b+\sqrt{b^2-4}}{2} \le \frac{-b + |b|}{2} = 0$

here, we arrive at the inequality of $A\le 0$ , which is impossible since $A=e^x >0$ , so b cannot be positive.

case 2: $b<0$

$\implies -b = |b| > 0$

if we take the positive strand, we will have two positive terms, $|b|$ and $\sqrt{b^2-4}$ adding together, hence A is positive, all is well.

if we take the negative strand, remembering that $|b|\ge \sqrt{b^2-4}$ , hence $\frac{|b|-\sqrt{b^2-4}}{2}\ge 0$ , it is also acceptable.

ALAS, $b\in (-\infty,-2]$

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