1. ## some exam questions

I've got a test on Monday and I have to get these off my back, thanks

Graphs
Find algebraically the implied domain of $\frac{1}{\sqrt{4x-x^2}}$. (3 marks)

7 lines to write your answer! Wut? I don't have a concrete method of doing this either... since are Not supposed to use calculus . But we are of course allowed limits. Besides... I've never seen this function before... we never studied it in class. I'd be happy if someone could also find the range, just so if another question asks about that then I'm prepared.

The rule for the inverse function of

$f:[1,\infty) \rightarrow R, f(x)=(x-1)^2+4$ is:

A $f^{-1}(x)=1+\sqrt{x-2}$

B $f^{-1}(x)=1-\sqrt{4-x}$

C $f^{-1}(x)=1+\sqrt{x-4}$

D $f^{-1}(x)=1-\sqrt{x-4}$

E $f^{-1}(x)=1+\sqrt{4-x}$

Hmm? I thought it wouldn't have an inverse function because of the plus-minus roots.

Probability
Bill, Bob, and Ben fire one shot at a target.

The probability that Bill hits the target is $\frac{4}{5}$

The probability that Bob hits the target is $\frac{3}{4}$

The probability that Ben hits the target is $\frac{2}{3}$

i) Find the probability that only Ben's shot hits the target. (2 marks)
ii) given that only one shot hits the target, it is the shot from Ben. (6 marks)

(What's the difference between these two questions??)

Ben enters a competition firing 3 shots at the same target. The probability
he hits the target remains constant at $\frac{2}{3}$. Given that he hits the target at least once, what is the probability that he hits it
exactly twice? (4 marks)

2. Originally Posted by DivideBy0
Graphs
Find algebraically the implied domain of $\frac{1}{\sqrt{4x-x^2}}$. (3 marks)

7 lines to write your answer! Wut? I don't have a concrete method of doing this either... since are Not supposed to use calculus . But we are of course allowed limits. Besides... I've never seen this function before... we never studied it in class. I'd be happy if someone could also find the range, just so if another question asks about that then I'm prepared.
You don't need Calculus for this. To find the domain, start by looking for places where the function doesn't exist. Such as where the denominator is 0 or where you get a negative under a square root.

Originally Posted by DivideBy0
The rule for the inverse function of
Originally Posted by DivideBy0

$f:[1,\infty) \rightarrow R, f(x)=(x-1)^2+4$ is:

A $f^{-1}(x)=1+\sqrt{x-2}$

B $f^{-1}(x)=1-\sqrt{4-x}$

C $f^{-1}(x)=1+\sqrt{x-4}$

D $f^{-1}(x)=1-\sqrt{x-4}$

E $f^{-1}(x)=1+\sqrt{4-x}$

Hmm? I thought it wouldn't have an inverse function because of the plus-minus roots.
Let $y = f(x) = (x - 1)^2 + 4$

Now switch the roles of x and y:
$x = (y - 1)^2 + 4$

Now solve for y:
$(y - 1)^2 = x - 2$

$y - 1 = \pm \sqrt{x - 2}$

Now, you are correct in that you can't have the $\pm$ and have the inverse function be an actual function. There are two things you can do at this point if you get a question like this on a test:
1) Call your instructor over and point out the problem. (S)He will tell you what to do about it.

2) Use the convention that, whenever you are taking the square root of a (well-defined) number (such as 45, not such as $x^2 - 35$) that we select the positive sign. Make sure to note that you are doing this on your paper.

So select the + sign on the square root. When you are finished solving for y, then $y = f^{-1}(x)$.

-Dan

3. Originally Posted by topsquark
You don't need Calculus for this. To find the domain, start by looking for places where the function doesn't exist. Such as where the denominator is 0 or where you get a negative under a square root.
-Dan
Thanks, I looked for $x:x^2-4x>0$ and got domain: $x\in (\infty, 0) \cup x\in (4,\infty)$.

But the range it seems is harder! How can I do that?

4. Originally Posted by DivideBy0
Thanks, I looked for $x:x^2-4x>0$ and got domain: $x\in (\infty, 0) \cup x\in (4,\infty)$.

But the range it seems is harder! How can I do that?
Hi,

I can't see topsquark around so I'll take over:

1. The radical of your function is: $4x-x^2$ (you changed the order of the terms)

$4x-x^2 >0~\iff~0

2. If x approaches zero or if x approaches 4 the difference approaches zero too thus the value of the function will increase to infinity. The greater the radical the smaller the value of the function. Obviously at x = 2 the radical has it's greatest value that means the function has it's smallest value:

$f(2) = \frac12$ Therefore the range is $\left[\frac12, \infty\right)$

5. Originally Posted by DivideBy0
I've got a test on Monday and I have to get these off my back, thanks
Graphs
Find algebraically the implied domain of $\frac{1}{\sqrt{4x-x^2}}$. (3 marks)
First find the domain $4x - x^2 > 0$ so $x^2 - 4x < 0 \implies (x-2)^2 < 4$ thus $0< x<4$.

Here is how to find the range algebraically*. But it has a lot of algebra in it (I have a lot of faith in you, I know you can handle it). What we are really asking is:
"Find all $y$ (real numbers) so that:
$\frac{1}{\sqrt{4x-x^2}} = y$
Has a solution in the domain, i.e. has a solution for $0."

So we need to solve this equation for $x$. But this equation is not always solvable for instance when $y\leq 0$ because the LHS is positive. So our task is to determine when it is possible to solve it for $0. So the first obersvation is that $y > 0$ otherwise it is impossible.

Begin by squaring,
$\frac{1}{4x-x^2} = y^2$
Thus,
$4x-x^2 = \frac{1}{y^2}$
Thus,
$x^2 - 4x + \frac{1}{y^2} = 0$
This is a quadradic equation it has real solutions when the discriminant is non-negative:
$16 - \frac{4}{y^2} \geq 0$
Thus,
$y \geq \frac{1}{2}$ (the other solution is ignored because we already established that $y>0$).

So it seems the range is $y\geq \frac{1}{2}$ is you want we can check this:

This means the solution is,
$x = \frac{4 \pm \sqrt{ 16 - \frac{4}{y^2} }}{2} = 2 \pm \sqrt{4 - \frac{1}{y^2}}$
But since we must have $0 < x < 4$:
$0 < 2 \pm \sqrt{4 - \frac{1}{y^2}} < 4$
Thus,
$- 2 < \pm \sqrt{4 - \frac{1}{y^2}} < 2$
Thus,
$\left| \pm \sqrt{4 - \frac{1}{y^2}} \right| < 2$
Thus,
$\sqrt{4 - \frac{1}{y^2}} < 2$
Which is of course true telling us that these values of $y$ give us a solution in the domain.

*)This is my own method which nobody apparently seems to use nor were taught in school.

6. Thanks so much!

7. Could someone check if this is right?

Consider $f(x)=\frac{1}{\sqrt{3x^2-4x}}$.

Domain:
$4x>3x^2 \implies 3x^2-4x<0\implies x(3x-4)<0$

$\implies x \in (0,\frac{4}{3})$

Range:
$y=\frac{1}{\sqrt{3x^2-4x}}$

$y^2=\frac{1}{3x^2-4x}$

$3x^2-4x-\frac{1}{y^2}=0$

$\Delta = 16+\frac{12}{y^2} >0$

$y^2 > -\frac{16}{12}$

Solving for y we get imaginaries, but looking back at the discriminant, this means y can actually take any value it likes without being 0, right?

$x=\frac{4\pm \sqrt{16+\frac{12}{y^2}}}{6}$

$x=\frac{2}{3} \pm \sqrt{4(4+\frac{3}{y^2})}$

$x=\frac{2}{3} \pm 2\sqrt{4+\frac{3}{y^2}}$

$0 <\frac{2}{3} \pm 2\sqrt{4+\frac{3}{y^2}}<\frac{4}{3}$

$-\frac{2}{3} < \pm 2\sqrt{4+\frac{3}{y^2}}<\frac{2}{3}$

$0<\left| \pm 2\sqrt{4+\frac{3}{y^2}}\right|<\frac{2}{3}$

$0<\sqrt{4+\frac{3}{y^2}}<\frac{2}{6}$

(Honestly, I still don't get why we have to check this. To me it seems almost independent to y. Is the first part of finding the range good enough for any problem?)

Domain: $x \in (0,\frac{4}{3})$

Range: $y \in R$ \ {0}

8. Ah, so you like my way of finding ranges.

$f(x) = \frac{1}{\sqrt{3x^2 - 4x}}$
We require that $3x^2 - 4x > 0$ so $x(3x - 4) > 0$. So (1) $x>0 \mbox{ and }3x - 4 > 0$ or (2) $x<0 \mbox{ and }3x - 4 < 0$.
In (1) the solution would be $x>0 \mbox{ and }x>4/3 \implies x>4/3$.
In (2) $x< 0 \mbox{ and }x<4/3 \implies x<0$.
Thus, $x>4/3 \mbox{ or }x<0 \implies \mbox{ domain } = (-\infty,0) \cup (4/3,\infty)$.
---
To find the range we need to solve,
$y = \frac{1}{\sqrt{3x^2 - 4x}}$
We should realize this means $y>0$ because otherwise RHS is positive.
Now take reciprocals and square,
$y^{-2} = 3x^2 - 4x$
$3x^2 - 4x - y^{-2} = 0$
To have a real solution we need,
$16 + 12 y^{-2} \geq 0$
Realize that this is always true because $y^{-2} \geq 0$ (in fact $>0$ because we said $y>0$) since it is a square.

And now check that this is a solution in the domain.*
So the range is $y> 0 \implies (0,\infty)$.

*)The reason why we need to check is because the solution that we get should be in the domain of the function. So if for say $y=2$ we get a solution $x=1$ then we have to disregard it because we are working on $(-\infty,0)\cup (4/3,\infty)$.

9. Yay I got 65/70 on the test! And it's actually meant to be for year 11 and I'm in year 10! Thanks for the help

10. Originally Posted by DivideBy0
Yay I got 65/70 on the test! And it's actually meant to be for year 11 and I'm in year 10! Thanks for the help
What did you get wrong? Was it some stupid mistake?

11. I'm getting the test back in 5 days, bleh. Tomorrow is VCE correction (holiday), then the weekend, then we have 2 days holiday for the Melbourne Cup.
I remember now I definitely lost two marks because I forgot to do a question... you see in tests I get pretty jumpy, as in, I jump all over the test paper, if a question turns me off I'll leave it for later. So basically I decided to leave this one for later, and since it was tucked in at the bottom of a page I didn't notice when I came back. Yeah, unforgivable.
If the math head thinks it's a good enough score then I'll take another test and if that is good enough then I'll be able to start the 'advanced' course next year.
As for the other 3 marks, I'll just have to wait and see...