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# another q i dont get! watch

1. p2 jan 04 edex.

4.

the function f is even and has domain R for x>=0
f(x)=x2 -4ax

sketch the curve for y=f(x) showing co-ords of crossing points.

my curve was a 2u" shape that crosses at (0,0) and (4a,0)

markscheme gives a "curved w" shape that crosses at (-4a,0) touches (0,0) at the middle of the w and crosses at (4a,0) why does it cross at the -4a?

f(-2a)

i did

(-2a)2 -4a(-2a)
=4a2 +8a2
=12a2

but correct answer is that f(-2a)=f(2a)=-4a2 [it also says even function what does this mean?)

finally it asks dor solutions when a=3

solving gives 0=(x-15)(x+3) correct so far

but they say the answer is +/-15, not 15 and -3

2. I've just done that paper and got exactly the same answer as you! Firstly, everything we did wrong was because of the 'even function' bit. The definition of an even function is one which has the property f(x)=-f(-x). I.e. the graph is symmetrical around the line x=0. (From page 28 of my P2 text book). Therefore, you get the 'w' shape as you reflect the graph of x being greater or equal to 0 in the x-axis.

The reason f(-2x) gives the same answer as f(2x) is for the same reason: if you look at the correct graph, the negative x values have exactly the same y values as the positive x values, hence f(-2x)=f(2x).

Lastly, x is not -3 because it states at the beginning that x has got to be greater or equal to 0. x is +15 and -15 because again you have to take into account the other half of the graph that has been reflected in the x-axis due to it being an even function. Looking at the graph, it is obvious that there are two solutions for when y is 45, because if you draw a line y=45 (or any line in the positive y quadrants), it crosses the graph twice: once when x is positive, and once when x is negative.

Hope this makes sense and is correct! Anyone else agree with my answer?
3. The function is given by
f(x) = 4x² - 4ax for x >= 0

but it is an even function, so f(x) = f(-x)

replacing x by -x in the first expression, we get

f(x) = 4x² + 4ax for x <= 0

This is the function to use for x <= 0, a different domain!

we can write this as,

f(x) =
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\left{

\begin{array}{clcr}

4x^2 - 4ax, & \mbox{for x \ge \ 0} \\

4x^2 + 4ax, & \mbox{for x \le \ 0} \\

\end{array}

You should be able to get the MS answers now.

Edit: I'm a bit late , so I agree with jen05
Attached Images

4. (Original post by Jen05)
I've just done that paper and got exactly the same answer as you! Firstly, everything we did wrong was because of the 'even function' bit. The definition of an even function is one which has the property f(x)=-f(-x). I.e. the graph is symmetrical around the line x=0. (From page 28 of my P2 text book). Therefore, you get the 'w' shape as you reflect the graph of x being greater or equal to 0 in the x-axis.

The reason f(-2x) gives the same answer as f(2x) is for the same reason: if you look at the correct graph, the negative x values have exactly the same y values as the positive x values, hence f(-2x)=f(2x).

Lastly, x is not -3 because it states at the beginning that x has got to be greater or equal to 0. x is +15 and -15 because again you have to take into account the other half of the graph that has been reflected in the x-axis due to it being an even function. Looking at the graph, it is obvious that there are two solutions for when y is 45, because if you draw a line y=45 (or any line in the positive y quadrants), it crosses the graph twice: once when x is positive, and once when x is negative.

Hope this makes sense and is correct! Anyone else agree with my answer?
Thanks a lot, you too fermat, still one thing i am confused though, how can it be -15 if x>0? and if x>0 why doesnt the graph just "stop" at the y axis?
Also how would i know it was an even function thingy from the question?
5. You know it is even, as it says in the question it is.

And how can it be -15?

Because although the equation for the function is only given for x=/> 0 it doesn't mean that that is where the function ends. As it is even you know that it is the same on the other side of the y axis (i.e. f(x) = f(-x) eg. cos(x)). So the domain of the function is not restricted to x =/> 0 onlye the part of the function with the given equation is.

I think that Fermat illustrates this well by writing out the whole function.
6. I think that Fermat illustrates this well by writing out the whole function.
I agree. I am actually a bit hazy on the mathematical approach, as we have never been taught even functions, and there is only about a paragraph on it in my text book, but no examples. However, luckily I think (hope!) I understand properly now (with hours to go, aaahhhh!). Good luck to everyone who has P2 tomorrow, I hope the examiners have been kind!

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