VCE Methods Question Thread!

[b^3]

Thank you b^3, that makes so much more sense   

Where do you find the modulus symbol | on a CAS calc?

Depends on which calc, but if you have the Ti-nspire (the newer one) then it should be under [Ctrl] + [=] I think, (I have the old grey one and think it's different, but going by this http://www.techpoweredmath.com/wp-content/uploads/2011/07/ti-nspire-cx.jpg , where the blue buttons is what you can get if you click ctrl first). You can also type in "abs(blah)" as well.

If you have the Casio Classpad, then someone else hopefully should be able to help.

[Snorlax]

Thank you b^3, that makes so much more sense   

Where do you find the modulus symbol | on a CAS calc?

On my CAS, I press that button on the left of the book (which is on top of the division symbol)
Then it'll be the first column, second row.

 

[SocialRhubarb]

Alternatively, you can use abs(    ), short for absolute value.

[Nato]

Hey guys,
i need to find the implied domain for the following



i haven't really encountered something like this before.
But I had a feeling i should somehow separate the two square root functions as if they were functions on their own, and find their individual  domain. is this in anyway correct?

I'm stuck beyond this point. Any help would be appreciated.
thanks

[b^3]

The domain of will be given by the intersection of the domains for and as has to satisfy both. If it satisfies one but not the other, then isn't defined, e.g. for , but is not defined on .

So for we have , so .
For we have .

Now the values of that satisfies both of those is then , so the domain of is , the intersection of the sets and .

Hope that helps!

EDIT: Somehow managed to use subsection instead of intersection, fixed.

[Nato]

The domain of will be given by the intersection of the domains for and as has to satisfy both. If it satisfies one but not the other, then isn't defined, e.g. for , but is not defined on .

So for we have , so .
For we have .

Now the values of that satisfies both of those is then , so the domain of is , the intersection of the sets and .

Hope that helps!

EDIT: Somehow managed to use subsection instead of intersection, fixed.

not only was that a fast reply, but really thorough, and helped me understand it.
thanks b^3

[Nato]

not only was that a fast reply, but really thorough, and helped me understand it.
thanks b^3

is there anyway we can find the range of that function?

[b^3]

is there anyway we can find the range of that function?

Depending on what you've covered so far. You'd substitute in the endpoints from the domain you've found to find their corresponding values, and then find the turning points and find their values and see what it ranges over. The calculus part you might not have done yet, (if you aren't meant to do it that way for this exercise then just graph it on cas and trace to find it, but you'll need to know the calculus part for later in the year anyways).

So differentiating and equating the derivative to zero gives.


Then finding the endpoints,
,
, .
, .

Our minimum is and our maximum is , so our range is , note the square brackets to include the endpoints.

[Nato]

any hints for this??

the implied domain for

[Zealous]

any hints for this??

the implied domain for

Simplify the fraction inside of the square root using long division. You will get a hyperbola.

We know a square root function is defined when the inside is equal or greater than 0. Once you simplify the fraction, you'll get a hyperbola, find the domain in which this is equal to or greater than 0 and you'll get the implied domain of the square root function.

Hopefully that makes sense, you asked for a hint =p

[b^3]

There are three things to note when trying to find implied domains.
 - If you have a square root, then anything under the square root has to be equal to or greater than zero (since we can't take the square root of a negative number in methods).
 - If you have a fraction, the denominator of the fraction can't be zero, as we can't divide by zero, the function is undefined at that point.
 - If you have a log, then whatever is inside the log has to be greater than zero, since we can't take the log of zero or a negative number.

There is also the partly trick case which some people fall for, where you have a square root in the denominator of a fraction, where for the square root you need , but to satisfy the fraction you need , and so need to take whats common to both, .
...Now I haven't actually got to the question specifically yet.

Use the rules above, but note what happens when you multiply the denominator of up. If it is positive then everything goes well, if it is negative then you need to flip the inequality sign, but taken into account the region where it is negative.

try it first before looking at the working here

So at the moment we have and , but we not take into account that the denominator cannot be zero, that is , so . This gives us our domain of

You could also use this little trick sometimes to help you out, but in this situation it just makes more work than is needed.

Spoiler

[Bluegirl]

Not doing maths for so long has made me brain dead
I need help with this:

If x^3 + 4x^2 -12x + 14 = x^3 + (mx+n)^2 + 5 for all values of x, find the values of m and n.

Thanks!

[Stick]

Not doing maths for so long has made me brain dead
I need help with this:

If x^3 + 4x^2 -12x + 14 = x^3 + (mx+n)^2 + 5 for all values of x, find the values of m and n.

Thanks!

I'm going to equate both sides.



(factorising the quadratic part of the expression)

(completing the square)

(completing the square and simplifying)

(expanding)

(incorporating the pronumeral into the bracket)

[b^3]

So we need to get what we have in the form , but at the moment we have instead of . So we'll split that up, then factorise the quadratic expression we have in the middle.

So we're trying to factorise .
Since we have a 4 in front of the term, our expression will be either of the form or . Since we are told we need to have it in the form , we will use the latter expression.
Now we need a pair of numbers that will result in using the above to add to and multiply to give .
So firstly lets try the pair. So factors of 9 are 9 and 1 or -9 and -1, 3 and 3 or -3 and -3. Again since we have a perfect square the two numbers need to be the same, and as we have negative 12, we need to use the two -3's to give us the middle term.
.
So we have , where and are and respectively.

EDIT: Stick has a different method above, either will work. Just choose the one you feel more comfortable with.

[Strawberrry]

Can anyone explain how to do f(x)= |x| - |x+1|-2  in terms of graphing? Is it better to add or find the difference? Thanks