Reviewing Potential Errors

[BAH0003]

Hi,

I was wondering if anyone is able to check over a major assignment I have completed, and point out any errors I may have made. I am fairly confident it is all correct, but if someone can give it another once over that would be great. If you're not able to do this, is it okay if you can check over 2e), 6c) and 5f) at least. The questions follow along with each other, so I have attached a link with each page of the assignment (it is six pages long) that will open in another browser.

With 2e) - I have written that it is reflected on the x-axis. Is this the correct terminology, or would it just be "on its axis"?

With 5f) - I am new to using derivatives so am not sure on this one.

With 6c) - I am concerned whether the answer is either 113.2m or 113.3m. I put 113.3m, but this is with using the previously rounded values in my distance formula. Would I have to use the original non-rounded values in it for it to be correct?

LINK TO ASSIGNMENT:

https://www.dropbox.com/s/tmb9qge907ac5w8/Maths%20Assignment.pdf?dl=0

Greatly Appreciated and Kind Regards ,
BAH0003

[AlphaZero]

Some of what I'll write isn't necessarily picking on things that are incorrect, but rather ways to improve some of your responses, or ways to be more efficient in your responses.

Question 2b
I understand what you're trying to say here but the term "narrower" isn't that precise. For example, takes the graphs of  \(y=\sqrt{1-x^2}\)  and  \(y=2\sqrt{1-x^2}\)  (you can sketch these yourself). The \(2\) actually has nothing to do with the narrowing of the graph since the 'width' of the graph stays the same. A more precise answer (and a simple one) is that when compared to the graph of  \(y=x^3\),  the \(2\) acts as a dilation from the \(x\)-axis. The idea of 'narrowing' actually requires a dilation from the \(y\)-axis, which in this case is technically possible with a factor of  \(2^{-1/3}\)  when comparing to the graph of  \(y=x^3\).  Your response isn't actually wrong, but could be a bit nicer


Questions 2c and 2d
It's great that you recognise how the point of inflection of the graph will move, but you don't need those sentences. Precise answers could be  "Translation of 4 in the positive \(y\)-direction"  and  "Translation of 10 units in the negative \(x\)-direction"  for these parts respectively.


Question 2e
Your answer is actually incorrect. A reflection in the \(x\)-axis would actually map the graph of  \(y=2(x-3)^3-5\)  to the graph of  \(y=-2(x-3)^3+5\),  which clearly isn't simply replacing the \(2\) with a \(-2\).  There are actually two simple transformations that occur:  a reflection in the \(x\)-axis, followed by a translation of 10 units in the negative \(y\)-direction.  You can graph these to check that replacing the \(2\) with a \(-2\) is not a simple reflection in the \(x\)-axis. As for terminology, reflection in the __-axis is fine.


Question 2f
Best way to disprove a statement is to provide a counterexample.  For example,  \(f(x)=x^3-x\)  is a cubic that is many-to-one since \(f(-1)=f(0)=f(1)=0\).


Question 3c
I would argue that a suitable domain is \((0,\,6)\) (not including 0 and 6), but I guess a case can be made for both answers.


Question 3d
The shape of your graph needs some fixing. I suggest you use your CAS to check the shape of the graph. Some obvious points to fix are:  the concavity the graph to left of the turning point should be concave down,  and the graph doesn't approach 0 gradient at \(x=6\) (it's not a double zero).


Question 3e
If you want, you can use calculus techniques here: \[\frac{dV}{dx}=12x^2-120x+216=0\implies x=5-\sqrt{7}\\
V(5-\sqrt{7})=56\sqrt{7}+80\ \text{ cm}^3\]

Question 4c
Make sure that you read the question carefully. It says to sketch the graph over a given domain and to label stationary points (which you haven't done). Graph shape needs some fixing too. Your turning points look like sharp points (they should be smooth) and it looks like they go down to the same \(y\)-level (which they don't).


Question 6a.ii
This answer isn't sufficient.  For example, the graph of  \(y=x(x-1)(x+1)\)  begins below the \(x\)-axis too but the coefficient of the \(x^3\) term isn't negative (you can check that it's \(1\)).  A more precise answer is that  \(y\to -\infty\)  as  \(x\to\infty\)  and so the coefficient of the highest order term \((x^4)\), which is \(d\) must be negative.


Question 6b.i
This isn't your fault, but technically we don't know whether \(a=-70,\,b=180\) or the other way around lol. Also, the values of \(a,b,d\) should be treated as dimensionless.


Question 6c
The wording "round all values to one decimal place" I think just refers to the final answer. It's not advisable to round during working since it leads to rounding errors, such as what has happened here. The correct answer is actually \(113.2\) metres, correct to one decimal place.