Author Topic: Silly Questions Thread (Read 33119 times) Tweet Share

the.watchman · « **Reply #195 on:** April 21, 2010, 05:25:26 pm »

Quote from: m@tty on April 21, 2010, 05:16:12 pm

One 'shortcut' is to let the $Ran_f =Dom_g$ and then work out $Ran_g$ on that domain, this range is the range of $g[f(x)]$ .

Exactly what I meant ... but expressed WAY neater

Aqualim · « **Reply #196 on:** April 22, 2010, 04:13:38 pm »

Simplify;
$2log_e(3x+2)-log_e(x)$

Could you go any futher than this;
$log_e(\frac{(3x+2)^2}{x})$

$\therefore 2log_e(\frac{(3x+2)}{x})$

/0 · « **Reply #197 on:** April 22, 2010, 04:20:52 pm »

The last step is incorrect,

$\log_e\left(\frac{(3x+2)^2}{x}\right) \neq \log_e\left[\left(\frac{3x+2}{x}\right)^2\right]$

I think your second last step is as far as you can go. Even the original equation I would say looks 'simplified'. It's a matter of opinion.

(3500 posts, yay! XD)

Aqualim · « **Reply #198 on:** April 22, 2010, 06:31:39 pm »

Also when dealing with matrices;

A reflection in the y-axis is different to a reflection from the y-axis yeah?

reflection from the y-axis = $\begin{bmatrix} -1 &0 \\ 0&1 \end{bmatrix}$

reflection in the y-axis = $\begin{bmatrix} 1 &0 \\ 0&-1 \end{bmatrix}$

If I'm correct?

Blakhitman · « **Reply #199 on:** April 22, 2010, 06:37:08 pm »

I haven't heard "from the whaterver axis" when it comes to reflections *scratch*.

Aqualim · « **Reply #200 on:** April 22, 2010, 06:43:36 pm »

actually don't worry, my mistake.

Aqualim · « **Reply #201 on:** June 30, 2010, 03:01:06 pm »

Would this is classified as a suitable line of working for a 3 mark question? My main concern is writing 'solve' in the line of working, which would be taken directly from the calculator.

6. Find in exact form all real solutions to the equation $2^{3x}+2^{2x}-2^{-x}=1$
$(2^x)^3+(2^x)^2-(2^x)^{-1}-1=0$

Let $2^x=a$

$solve(a^3+a^2-a^{-1}-1=0,a)$

$a=-1~or~a=1$

Since $2^x=a$

$2^x=1~or~2^x=-1$

$x=undef.~or~x=0$

$\therefore~x=0$

TrueTears · « **Reply #202 on:** June 30, 2010, 03:07:19 pm »

you can definitely solve that by hand so I'd say you need to show working by hand.

brightsky · « **Reply #203 on:** June 30, 2010, 03:09:48 pm »

Let $2^x = a$

$a^3 + a^2 - a^{-1} - 1 = 0$

Multiply a to both sides:

$a^4 + a^3 - 1 - a = 0$

$a^3(a + 1) - (a + 1) = 0$

$(a^3 - 1)(a+1) = 0$

$(a-1) (a^2 + a + 1) ( a+1) = 0$

By the null factor law,

$a - 1 = 0$ or $a^2 + a + 1 = 0$ or $a + 1 = 0$

$a = 1$ or "a has no real solutions as $\Delta < 0$ " or $a = - 1$

Hence $a = 1$ or $a = -1$

$\because 2^x = a$

$\therefore 2^x = 1$ or $2^x = -1$

$x = 0$ or "x has no solutions"

Hence $x = 0$ is the only solution.

Blakhitman · « **Reply #204 on:** June 30, 2010, 03:10:23 pm »

In further apparently, writing the solve part (or any calc functions) is sufficient working out

.

Aqualim · « **Reply #205 on:** June 30, 2010, 04:40:46 pm »

For the function $f:(-\infty , 0] \to R$ with the rule $f(t)=\frac{3t}{t^2+1}$ find the rule for the inverse function.

When finding the rule, would I have to use the Quadratic Formula at some point?

the.watchman · « **Reply #206 on:** June 30, 2010, 05:00:33 pm »

Quote from: Aqualim on June 30, 2010, 04:40:46 pm

For the function $f:(-\infty , 0] \to R$ with the rule $f(t)=\frac{3t}{t^2+1}$ find the rule for the inverse function.

When finding the rule, would I have to use the Quadratic Formula at some point?

Absolutement! Unless you want to complete the square manually

Aqualim · « **Reply #207 on:** June 30, 2010, 05:03:20 pm »

Quote from: the.watchman on June 30, 2010, 05:00:33 pm

Quote from: Aqualim on June 30, 2010, 04:40:46 pm
For the function $f:(-\infty , 0] \to R$ with the rule $f(t)=\frac{3t}{t^2+1}$ find the rule for the inverse function.

When finding the rule, would I have to use the Quadratic Formula at some point?

Absolutement! Unless you want to complete the square manually

How would you do it manually?

the.watchman · « **Reply #208 on:** June 30, 2010, 05:04:54 pm »

Quote from: Aqualim on June 30, 2010, 05:03:20 pm

Quote from: the.watchman on June 30, 2010, 05:00:33 pm
Quote from: Aqualim on June 30, 2010, 04:40:46 pm
For the function $f:(-\infty , 0] \to R$ with the rule $f(t)=\frac{3t}{t^2+1}$ find the rule for the inverse function.

When finding the rule, would I have to use the Quadratic Formula at some point?

Absolutement! Unless you want to complete the square manually

How would you do it manually?

By completing the square...

Let $t=f(y)$

$t=\frac{3y}{y^2+1}$

$ty^2 -3y+t=0$

Then complete the square like normal

Aqualim · « **Reply #209 on:** July 01, 2010, 01:28:45 pm »

Quote from: the.watchman on June 30, 2010, 05:04:54 pm

Quote from: Aqualim on June 30, 2010, 05:03:20 pm
Quote from: the.watchman on June 30, 2010, 05:00:33 pm
Quote from: Aqualim on June 30, 2010, 04:40:46 pm
For the function $f:(-\infty , 0] \to R$ with the rule $f(t)=\frac{3t}{t^2+1}$ find the rule for the inverse function.

When finding the rule, would I have to use the Quadratic Formula at some point?

Absolutement! Unless you want to complete the square manually

How would you do it manually?

By completing the square...

Let $t=f(y)$

$t=\frac{3y}{y^2+1}$

$ty^2 -3y+t=0$

Then complete the square like normal

Ok I got two answers;
$\frac{3+\sqrt{9-4t^2}}{2t}~and~\frac{3-\sqrt{9-4t^2}}{2t}$

$\therefore \frac{3-\sqrt{9-4t^2}}{2t}$
since the domain says its negative only, then I'm assuming the inverse would be quadratic with the negative in it. But after sketching on the calculator it doesn't look like its inverse?

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