VCE Methods Question Thread!

[Evolio]

So if e is to the power of x to the power of a fraction,then you take the whole thing to the front?

[aspiringantelope]

You need to put a multiplication symbol between letters. Your CAS calculator treats the input  \(\texttt{ax}\)  as a single variable instead of the intended  \(a\cdot x\).  This would be the reason why you were getting  \(\texttt{false}\)  as your output. In general, you should leave the syntax of your input as is. Some other ways of writing simultaneous equations are understood by the software, but to be honest, it would be a bit too time consuming trying to figure out what inputs work rather than just to use the one provided when you press  \(\texttt{Menu}\to \texttt{3}\to \texttt{7}\to \texttt{1}\).  (Besides, I'm not sure why you would want to remove the curly brackets since they're placed there for you anyway).

Edit: if you only have one variable to solve for, curly brackets are not required. For example:
\[\texttt{solve}\Big(x^2-1=0,x\Big)\]Yes, it is:  \(1\pm\dfrac{1}{\sqrt{2}}=1\pm \sqrt{\dfrac12}\).  The former does look a little more pleasing though.
Show that the discriminant of the resulting quadratic is greater than zero.

Ok! Thanks a lot! For the curly brackets
I actually didn't read before and didn't go from Menu - 3 - 7 - 1 but instead typed it all in manually.

And for the discriminant one I'll try it soon!

Thanks again @AlphaZero

[Evolio]

Thank you MB_ and Alpha Zero!
I guess I got confused between the two.
All good now.

[Evolio]

Hi.
I was wondering how you would differentiate this?

[DBA-144]

Hi.
I was wondering how you would differentiate this?

Please avoid double posting if you are asking questions/posting,etc. within such a short time span. (I'm no mod, but of course it helps keep the thread more organised for everyone)

I don't know how to use latex, but here's how you do it:
1. Use the substitution u=cos 2x-pi/4. Then it becomes u^3. then differentiate using the chain rule. Make sure that you also differentiate the cos (2x -pi/4) too.

That is to say, your final answer should look like the following:

-2  x 3 cos^2(2x-pi/4) sin (2x-pi/4). (deliberately didn't simplify to show that you get the negative from diffing cos to sin, and the 2 from the chain rule from the same thing and then the 3 from the substitution you made.

You are essentially using the chain rule twice. Don't get scared by the term, it's not that scary once you take some time to think about it.

Hope this helps.

[Evolio]

Please avoid double posting if you are asking questions/posting,etc. within such a short time span. (I'm no mod, but of course it helps keep the thread more organised for everyone)

I don't know how to use latex, but here's how you do it:
1. Use the substitution u=cos 2x-pi/4. Then it becomes u^3. then differentiate using the chain rule. Make sure that you also differentiate the cos (2x -pi/4) too.

That is to say, your final answer should look like the following:

-2  x 3 cos^2(2x-pi/4) sin (2x-pi/4). (deliberately didn't simplify to show that you get the negative from diffing cos to sin, and the 2 from the chain rule from the same thing and then the 3 from the substitution you made.

You are essentially using the chain rule twice. Don't get scared by the term, it's not that scary once you take some time to think about it.

Hope this helps.

Yeah, I got the same answer as you but the answer is saying something weirdly different. Is the answer wrong?

[DBA-144]

Yeah, I got the same answer as you but the answer is saying something weirdly different. Is the answer wrong?

What's the answer?

[aspiringantelope]

Show that the discriminant of the resulting quadratic is greater than zero.

Hey! I've done this but when discriminant is greater than zero, it just shows that the parabola has two x-intercepts, not how many intersections.
How can I correlate with the y = x + 2 straight line so that I can show there are two points of intersection?

Thanks

[Evolio]

What's the answer?

[AlphaZero]

Hey! I've done this but when discriminant is greater than zero, it just shows that the parabola has two x-intercepts, not how many intersections.
How can I correlate with the y = x + 2 straight line so that I can show there are two points of intersection?

Thanks

You need to show that the discriminant of the resulting quadratic (the one obtained by setting  \(x+2=2x^2+6x-1\) ) to be greater than zero.

Proof

\[x+2=2x^2+6x-1\implies 2x^2+5x-3=0\] \begin{align*}\Delta&=25-4\!\times\!2\!\times\!(-3)\\
&=25+24\\
&=49\\
&>0\end{align*} Hence, the graphs of  \(y=x+2\)  and  \(y=2x^2+6x-1\)  intersect twice.

Yeah, I got the same answer as you but the answer is saying something weirdly different. Is the answer wrong?

This is a classic case of a textbook being really stupid. The answer provided is correct, but no student would obtain this by hand. The question was clearly placed into a calculator, and as it is programmed to simplify expressions as much as possible, it did some algebraic manipulation to remove the negative sign from the front of the expression.  This was done by recognising that  \(\sin\left(2x-\dfrac{\pi}{4}\right)=-\cos\left(2x+\dfrac{\pi}{4}\right)\),  which is completely useless for this question.

How it should be done by hand

\begin{align*}\frac{d}{dx}\left[\cos^3\left(2x-\frac{\pi}{4}\right)\right]&=3\cos^2\left(2x-\frac{\pi}{4}\right)\!\times\!\frac{d}{dx}\left[\cos\left(2x-\frac{\pi}{4}\right)\right]\\
&=3\cos^2\left(2x-\frac{\pi}{4}\right)\!\times(-2)\sin\left(2x-\frac{\pi}{4}\right)\\
&=-6\cos^2\left(2x-\frac{\pi}{4}\right)\sin\left(2x-\frac{\pi}{4}\right)\end{align*}

[DBA-144]

could you show how or why that is the case (why that simplification works)?

[AlphaZero]

could you show how or why that is the case (why that simplification works)?

\begin{align*}\sin\left(2x-\dfrac{\pi}{4}\right)&=\sin\left[-\left(\frac{\pi}{2}-\left(2x+\frac{\pi}{4}\right)\right)\right]\\
&=-\sin\left[\frac{\pi}{2}-\left(2x+\frac{\pi}{4}\right)\right] \tag{sine is an odd function}\\
&=-\cos\left(2x+\frac{\pi}{4}\right) \tag{complementary angle identity} \end{align*} A similar argument can be made for  \(\cos\left(2x-\dfrac{\pi}{4}\right)=\sin\left(2x+\dfrac{\pi}{4}\right)\)  (noting that cosine is an even function).

Substituting the results into the 'by-hand' expression for the derivative yields the calculator result.

[persistent_insomniac]

When describing transformations such as (attatched), does the order matter? I got dilate by 1/2 from x, reflect in y, reflect in x, right 4 and up 3 but the book says reflect in y, right 4, reflect in x, dilate by 1/2 from x and up 15/2.
Would my answer also be correct/acceptable?

[AlphaZero]

When describing transformations such as (attatched), does the order matter? I got dilate by 1/2 from x, reflect in y, reflect in x, right 4 and up 3 but the book says reflect in y, right 4, reflect in x, dilate by 1/2 from x and up 15/2.
Would my answer also be correct/acceptable?

Yes, the order in which you apply transformations matters, but note that you can adjust your transformations so that they are equivalent.
For example, the following sequences of the transformations are not equivalent
1.  dilation by factor 2 from \(x\)-axis, followed by a translation of 2 units in the positive \(y\)-direction
2.  translation of 2 units in the positive \(y\)-direction, followed by a dilation by factor 2 from the \(x\)-axis.

The first sequence maps the graph of some  \(y=f(x)\)  onto the graph of  \(y_1=2\,f(x)+2\).
The second sequence maps the graph of some  \(y=f(x)\)  onto the graph of  \(y_2=2\big(f(x)+2\big)=2\,f(x)+4\).
The convention is to give dilations and reflections first, and then translations, but of course, it is possible to produce an equivalent sequence of transformations that start with giving translations first, and then dilations and reflections. For example, equivalent to sequence 1 above is
3.  translation of 1 unit in the positive \(y\)-direction, followed by a dilation by factor 2 from the \(x\)-axis.
Your sequence of transformations neglects to consider the constant at the end of the first function. Indeed, it is appropriate to start with a dilation by factor \(1/2\) from the \(x\)-axis to remove the \(2\) in front of the square root, but you cannot forget to also multiply the \(3\) by the \(1/2\) as well. In other words \[y=2\sqrt{4-x}+3\quad\overset{1/2\text{ from }x\text{-axis}}{\longrightarrow}\quad y_1=\sqrt{4-x}+\frac32,\] and so clearly, your translation of \(3\) units up later no longer works.
I've actually answered this exact question a couple of weeks ago. Click here for a full explanation

[persistent_insomniac]

Yes, the order in which you apply transformations matters, but note that you can adjust your transformations so that they are equivalent.
For example, the following sequences of the transformations are not equivalent
1.  dilation by factor 2 from \(x\)-axis, followed by a translation of 2 units in the positive \(y\)-direction
2.  translation of 2 units in the positive \(y\)-direction, followed by a dilation by factor 2 from the \(x\)-axis.

The first sequence maps the graph of some  \(y=f(x)\)  onto the graph of  \(y_1=2\,f(x)+2\).
The second sequence maps the graph of some  \(y=f(x)\)  onto the graph of  \(y_2=2\big(f(x)+2\big)=2\,f(x)+4\).
The convention is to give dilations and reflections first, and then translations, but of course, it is possible to produce an equivalent sequence of transformations that start with giving translations first, and then dilations and reflections. For example, equivalent to sequence 1 above is
3.  translation of 1 unit in the positive \(y\)-direction, followed by a dilation by factor 2 from the \(x\)-axis.
Your sequence of transformations neglects to consider the constant at the end of the first function. Indeed, it is appropriate to start with a dilation by factor \(1/2\) from the \(x\)-axis to remove the \(2\) in front of the square root, but you cannot forget to also multiply the \(3\) by the \(1/2\) as well. In other words \[y=2\sqrt{4-x}+3\quad\overset{1/2\text{ from }x\text{-axis}}{\longrightarrow}\quad y_1=\sqrt{4-x}+\frac32,\] and so clearly, your translation of \(3\) units up later no longer works.
I've actually answered this exact question a couple of weeks ago. Click here for a full explanation

So does that mean dilation by 1/2 from x, reflection in y, reflection in x, right 4, up 15/2 is incorrect b/c the order is incorrect? I'm still a bit confused because when I apply these transformations I still get to the final function.