## Friday, November 20, 2015

### Analysis: Quiz 15

Question 1: Which of the following represents the graph of $y=\sqrt{2}x$?
A.
B.
C.

This is a very straightforward problem as if you could recognize $y=\sqrt{2}x$ is a linear function, then we can eliminate the graphs of B and C as they are curves instead of straight line and so this suggests A is the answer.

Question 2: Which of the following represents the graph of $y=\sqrt{x^2+1}$?
A.
B.
C.

This is also an easy question since $y=x^2+1$ is always concave up for all $x\in R$, so its square will also a concave up curve and we can cross the option C out and we need to figure out if A or B is the correct curve. We could do this by replacing $x$ with any numerical value and see which curve represents $y=\sqrt{x^2+1}$.

It makes thing easier if we pick $x=1$ as it would lead to $y=\sqrt{2}\approx 1.414$ so B must be the answer.

Question 3: Which of the following represents the graph of $y=\sqrt{2}x-\sqrt{x^2+1}$?
A.
B.

When we have a straight line function $y=\sqrt{2}x$ and another curve function $y=\sqrt{x^2+1}$, and we are required to subtract the second by the first, the resulting function will always be a curve. So A is the correct answer.

Question 4: Which of the following represents the graph of $y=\dfrac{\sqrt{2}\ln x}{2}$?
A.
B.

Note that we're given the graph of $y=\ln x$ so it would act as a reference to differentiate between A and B.

We know in order to get the graph of $y=\dfrac{\sqrt{2}\ln x}{2}$ from $y=\ln x$ , we have to vertically shrink $y=\ln x$  by a factor of $\dfrac{\sqrt{2}}{2}\lt 1$, and so the correct answer to this question is B.

Question 5: What can you observe and then conclude from the graphs A and B as shown below?
Can you tell, on the interval $0\lt x\lt 1$, which of them is greater?
A.
B.

The reason why the problem has specified to compare which of A and B is greater at the interval $0\lt x\lt 1$ is because the question setter wanted to assess students' real understanding in determining which two negative values is greater.

Of course any function that lies above in Cartesian diagram is greater than any other function that lies below it. Since A lies above B at the interval $0\lt x\lt 1$, so it is safe to say A is greater than B in that entire specified domain. In fact, A is always greater than B for all $x\ge 0$.

Question 6: Based on your answer in previous question, which of the inequality holds true for $0\lt x\lt 1$?
A. $\sqrt{2}x-\sqrt{x^2+1}\lt \dfrac{\sqrt{2}\ln x}{2}$

B. $\sqrt{2}x-\sqrt{x^2+1}\gt \dfrac{\sqrt{2}\ln x}{2}$

Since A represents $y=\sqrt{2}x-\sqrt{x^2+1}$ and B represents $y=\dfrac{\sqrt{2}\ln x}{2}$, we can easily conclude $\sqrt{2}x-\sqrt{x^2+1}\gt \dfrac{\sqrt{2}\ln x}{2}$ is correct.

Question 7: What representation of approach would you adopt if you're asked to prove whether $\sqrt{2}x-\sqrt{x^2+1}$ is greater than (or equal to) or less than (or equal to) $\dfrac{\sqrt{2}\ln x}{2}$ for $x\gt 0$?

A. Graphing approach.
B. Differentiation approach.
C. AM-GM inequality formula.

If the question is asked us to prove which of $\sqrt{2}x-\sqrt{x^2+1}$ versus $\dfrac{\sqrt{2}\ln x}{2}$ is greater at the interval $0\lt x\lt 1$, we could of course do it with graphing approach.

But we're now asked to prove the problem for $x\gt 0$, graphical approach is out of the question since we don't know the area behind $x=1$, the function of $y=\sqrt{2}x-\sqrt{x^2+1}$ always lies above $y=\dfrac{\sqrt{2}\ln x}{2}$ or if they will intersect at some point.

The safest ticket to prove which is greater is by using the differentiation approach. So B is the answer.

Question 8: What is the first derivative of $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}$?

A. $\sqrt{2}-\dfrac{2x}{\sqrt{x^2+1}}-\dfrac{1}{\sqrt{2}x}$

B. $\sqrt{2}-\dfrac{x}{\sqrt{x^2+1}}-\dfrac{1}{\sqrt{2}x}$

C. $\sqrt{2}-\dfrac{x}{\sqrt{x^2+1}}-\dfrac{2}{\sqrt{2}x}$

This is to test students' overall performance on differentiating using the chain rules of differentiation and B is the answer.

Question 9: You can justify the result $\sqrt{2}-\dfrac{2x}{\sqrt{x^2+1}}\gt \dfrac{1}{\sqrt{2}x}$ if and only if you have proved:

A. The maximum of $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}$ is greater than or equal to zero for all $x\gt 0$.

B. The minimum of $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}$ is greater than or equal to zero for all $x\gt 0$.

Since we know $\sqrt{2}x-\sqrt{x^2+1}\ge \dfrac{\sqrt{2}\ln x}{2}$, (equality occurs at $x=1$) that means if we can prove $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}$ has a minimum point and that minimum point occurs either at $0$ or somewhere above $0$, then we can conclude it will only touch or will never cross the x-axis, that means $f(x)\ge 0$ (equality occurs if the minimum touches at x-axis) so $\sqrt{2}-\dfrac{2x}{\sqrt{x^2+1}}\gt \dfrac{1}{\sqrt{2}x}$ must be true.

Question 10: If $x,\,y,\,z$ are positive real such that $xyz=1$, what operation would you do to the inequalities below to generate another inequlaity that holds and satisfy the given condition?

$\sqrt{2}x-\sqrt{x^2+1}\ge \dfrac{\sqrt{2}\ln x}{2}$

$\sqrt{2}y-\sqrt{y^2+1}\ge \dfrac{\sqrt{2}\ln y}{2}$

$\sqrt{2}z-\sqrt{z^2+1}\ge \dfrac{\sqrt{2}\ln z}{2}$

A. Multiply the three inequalities.
B. Subtract the third by the second and then subtract the resulting by the first.

We can make use of the info $xyz=1$ if we're to add the three inequalities since
\begin{align*}\dfrac{\sqrt{2}\ln x}{2}+\dfrac{\sqrt{2}\ln y}{2}+\dfrac{\sqrt{2}\ln z}{2}&=\dfrac{\sqrt{2}}{2}\left(\ln x+\ln y+\ln z\right)\\&=\dfrac{\sqrt{2}}{2}\left(\ln xyz\right)\\&=\dfrac{\sqrt{2}}{2}\left(\ln 1\right)\\&=0\end{align*}
So the summation ends up with $\sqrt{2}x+\sqrt{2}y+\sqrt{2}z-\sqrt{x^2+1}-\sqrt{y^2+1}-\sqrt{z^2+1}\ge 0$, or
$\sqrt{2}(x+y+z)\ge \sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}$.