A great many math questions have stumped me. So, I am going to expand the scope of this one. A great math question should satisfy 3 conditions 1) It should be very attemptable and tempting for a wide audience. It cannot be something relevant only to Math PhDs 2) It should stump the smart ones 3) It should have an aha moment. The solution should be so elegant that we hear a collective sense of joy from everyone assembled when the solution is mentioned.

Back in the late 90’s when I had pretensions of taking Math up seriously, I got the chance to attend one of those Olympiad camps in South India. A set of gifted math teachers would bring together 60 of the top 500 or so math kids from classes X and XI and prep them for the INMO. The batch I had the great misfortune of sitting with had some guys who went on to get medals at the IMO (which is a seriously big deal). The camp was a seriously chastening experience.

There was this fabulously enthusiastic 60-something whose name I think was Prof Visweshwaran who had a pathological smiling syndrome. He simply could not quit grinning from ear-to-ear at the obvious pleasure some of these math questions were giving him.

He started the session with a question borrowed from one of the other Olympiads. He said “Every natural number in the world has a multiple that has all the digits from 0 to 9 appearing at least once in it.” We all nodded as if this were the most obvious thing ever. He paused and proceeded with “Now, prove that.” He pushed, prodded, gave examples and taunted the kids for a good 5–6 minutes, all with barely suppressed glee writ all across his face.

This is a delicious question. It seems true, you can feel it, but is perhaps not that easy to prove. For instance, how do you prove that the number 3691 has a multiple that has the digits from 0 to 9 appearing at least once in it. We can sense that there should be some multiple that works, but how the hell does one prove it?

After another 5 minutes of aggressive prowling and glorious smiling, he gave us the solution. Write down the number 1234567890 on your notebook. Now whatever number you are given, divide 1234567890 with that number. For instance, if the number you are thinking of is 3691. Do a long-division of 1234567890 by 3691, you are obviously going to get some remainder. Now, just add whatever digits you need to add at the end of 1234567890 in order to make that number a multiple of 3691. For instance, a multiple of 3691 that has all the digits would be 1234567890909. This might not be the smallest, or most beautiful or the best. But it satisfies all the conditions.

The good Professor then proceeded to laugh and smile for another 3 minutes. I remember this idea like it was taught yesterday. Hope it gives you guys also an “Aha” moment.

If you’re doing mathematics research, then you must be tackling a maths problem. Sometimes, problems solve themselves relatively quickly. Other times, they take years.

I’ve recently turned my attention on the Polynomial Reconstruction Problem, which was set in 1973. (That’s before I was born.)

Is it true that, for [math]n>2[/math], the characteristic polynomial of a graph [math]G[/math] on [math]n[/math] vertices is determined uniquely by the collection of characteristic polynomials of vertex-deleted subgraphs of [math]G[/math]?

This problem remains unsolved to this day. My PhD supervisor used to work on this problem as well, and, at one point, thought she had a proof that the answer is yes. (Interestingly, there are more people who think that the answer is ‘no’, but counterexamples have never been found.)

My contribution to this problem is very minimal, at this point, but my latest papers, though they seem to have nothing to do with it, actually arose while I was trying my hand at this problem. I’m still far, far away from solving it, probably further away than many other people, and it may turn out that the ‘no’ people are indeed correct.

Being stumped by a problem is what research is all about.

You are correct , partially.

From your solution, I see that the part where you went wrong was the second step.You subtracted 1 year from murtaza's age , but you forgot to subtract 1 year from sara ‘ s age. The question says that last year murtaza was 4 times Sara's age , so you have to subtract 1 from both the ages.

Now, Understand that both your teacher and you are wrong. How? Well if you take sara ‘s age to be 3 then murtaza age will be 12. One year before sara age was 2 and murtaza age was 11.

So this is not the correct answer. It should be solved like this.

m=s+9

One year before sara's age =s-1

Thus, m-1=4 (s-1)

Notice that the whole of s-1 will be multiplied with 4, and not just ‘ s.’

m-1=4s-4

m=4s-3

4s-3=s+9

3s=12

s=4

So that's the solution. The solution given by the checker is wrong . Your logic is right. You just didn't know how to apply that.

I think the next number in this sequence is 88.

In this sequence, if you have an initial number, you can obtain the next number by grabbing the digits of the product of the two first digits of the initial number and then joining those digits with the digits of the sum of the first and last digits of the initial number.

For example, in this case the sequence starts with 492.
If we apply the method previously mentioned, we have that the next number would have to be expressed as:
the digits of 4 (first digit) * 9 (second digit)
joined with the digits of 4 + 2 (last digit)
4 * 9 = 36
4 + 2 = 6
366.

This goes on until the last known number in the sequence, so, in order to find the unknown number, we’ll have to apply the same method as before.
This number would have to be expressed as follows:
the digits of 8 (first digit) * 1 (second digit)
joined with the digits of 8 + 0 (last digit)
8 * 1 = 8
8 + 0 = 8
This brings us to the final answer of 88.

It is easiest to reason about what you know and don't know if you can draw a picture--- in this case, a Venn diagram. You know the size of the set of people who like Pure, who like Applied, and who like Statistics, though these may overlap. You know the size of Pure AND Applied, and you know the size of Pure AND Statistics. You also know the size of Pure AND Applied AND Statistics.

The final input given is a little hard to represent directly on the graph, we know that the union of (Pure and Applied) with (Pure and Statistics) and (Applied and Statistics) is 20.

Now a basic rule we can use over and over again is that when [math]A = B \cup C[/math], and [math]B[/math] and [math]C[/math] do not intersect, then the size of [math]A[/math] is the sum of sizes of [math]B[/math] and [math]C[/math]. In particular, if we know the size of (X and Y) and the size of (X and NOT y), then we can sum them to get the size of X.

So, in the digram above, we know that (Pure and Statistics) is 10, and (Pure and Statistics) and Applied is 8, so (Pure and Statistics) and NOT Applied, which I've labelled "z", must be 2.

The same rule helps us calculate x, and then the last condition in the problem gives us y as well.

Once we have x, y, and z filled in on the diagram, we can fill in the unlabelled components (those who like only one subject.) For example, the number of people who like only pure mathematics is [math]19 - 8 -x - z[/math].

One the diagram is complete (each intersection of Pure, Applied, and Statistics is filled in with the number of students), then the answers to the questions can simply be read off. The last question refers to students who do not fit into any of these subsets, so we have to use the very first piece of information--- that the universe of discourse consists of 50 students.

Trying to prove Napolean’s Theorem.

The theorem states that if you draw any triangle, then draw equilateral triangles on the sides of that triangle (either all inwards or all outwards), the centres of those equilateral triangles form an equilateral triangle. Here’s a picture to demonstrate such a construction:

I floundered on this proof so badly! The solution that I finally stumbled upon uses symmetry. You could prove it using by trig bashing or using complex numbers.

It’s been a while, so this might not be the way I solved it exactly. Either way, our first step is to add another equilateral triangle [math]\Delta BVU[/math]:

where the triangle [math]\Delta A_1HG[/math]is equilateral and shares a side with [math]\Delta GHI[/math]. We wish to show that [math]\Delta GHI[/math] is equilateral. This looks like

With this bit of insight, we can show that [math]\Delta GHI[/math] is equilateral with very little work. Note that if we rotate our configuration [math]\frac{2\pi}{3}[/math] around point [math]H[/math], the point [math]I[/math] is now where point [math]A_1[/math] used to be. Thus, the line segments[math] HI[/math] and [math]A_1H[/math] are the same length and are [math]\frac{2\pi}{3} [/math]apart. Similarly, if we rotate the figure clockwise through an angle [math]\frac{2\pi}{3}[/math] about the point [math]G[/math], the triangle centred at I moves again to the position of the [math]\Delta BKL[/math]. This tells us that [math]GI[/math] and [math]GH[/math] are of equal length and make an angle of [math]\frac{2\pi}{3}[/math]. Consequently, [math]GH[/math] bisects [math]\angle IGA_1[/math] and [math]\angle A_1HI[/math]. This means [math]\angle IGA_1[/math] and [math]\angle A_1HI[/math] are [math]\frac{\pi}{3}. [/math]Thus, [math]\Delta GHI[/math] is equilateral.

The solution to this problem is not trivial; the tricky step in this proof is finding the idea of drawing [math]\Delta BVU. [/math] When you find that idea, however, the whole problem falls into place. Isn’t geometry really beautiful?

All my images were made on GeoGebra (sorry they aren’t that high quality).

1. Note that 2a must be an integer so a = k/2 for some integer k. Note also that 2a > b so k > b.
2. Draw an isosceles triangle. Label the base b and the legs k/2. Find the perimeter.
3. Get your good friend Pythagoras to help you find the height of your triangle.
4. Find the area of your triangle. Square it. Set that equal to the perimeter. Clear fractions (optional).
5. You should now have a mildly scary looking equation involving bs and ks. Note that it is quadratic in k. Use the quadratic formula to solve for k in terms of b. If you have not made any mistakes up to here, a minor mathematical miracle will occur: the discriminant is a perfect square. Simplify as much as you can and discard the negative root, I got k = b + 16/b^2 but don’t take my word for it; do the math. It’s fun.
6. Figure out which whole number values of b will give whole number values for k. Check that k > b. The list is not too long. Divide each k by 2 to get a and you’re there.

The infamous thought experiment Schrodinger's Cat was never intended to be answered.

Schrodinger crafted the problem to point out what he saw as a flaw in the Copenhagen interpretation of quantum mechanics if it were hypothetically applied to larger scaled objects. The Copenhagen model states that objects on the quantum level do not settle into a definite state until observed. This means that before it is observed, the object can be treated as existing in a state of superposition, or all possible positions simultaneously.

The problem of the cat locked in a box, whose life depended on whether an atom admitted radiation, was intended to be absurd. It is absurd to say that until the box is opened, the cat is both alive and dead. This is one of the most misunderstood problems in pop culture as relates to physics, if not the most misunderstood.

Schrodinger was not using metaphor. He quite literally meant that it was impossible for a cat to be alive and dead simultaneously. He was not speaking to the quantum truth of superpositions, but he had engaged with Einstein in a series of letters about creating large scale structures which could hypothetically have these same properties. Einstein believed they could be created. Schrodinger did not.

So, please, stop saying that the whole point is that the cat is alive AND dead. It is not the point. The point is that the cat’s fate is determined even if no one ever opens the box because on the non-quantum level, observing an object does not determine its state. The cat is not a quantum particle. It can not be both alive and dead.

Taking you (Y) and your friend (F) example:

Currently you are 17 and he is 65 which makes the first equation as:

F = Y + 48

Now a year ago, your friend (64) was 4 times your age (16) which makes the second equation:

F - 1 = 4 (Y - 1)

Do you see what you missed?

When you say 1 year ago that means 1 year ago for both you and your friend.

Note: The solution in red doesnt seem correct to me. It should be

M - 1 = 4 (S - 1)

according to above logic. And it should solve down like this:

[math]M = S + 9[/math] eq. 1

[math]M - 1 = 4 (S - 1)[/math]

[math]M - 1 = 4S - 4[/math]

[math]M = 4S - 3[/math] eq. 2

Comparing Eq. 1 and Eq. 2

[math]S + 9 = 4S - 3[/math]

[math]12 = 3S[/math]

[math]S = 4[/math]

[math]M = S + 9[/math] from eq. 1

[math]M = 4 + 9 = 13[/math]

Now you can check:

M is 13 years old. S is 4 years old. M is 9 years older than S.

One year ago, that is when M was 12 and S was 3, M was 4 times the age of S.

Hope it helps!

If you're able to picture it mentally, you'll save time, which is valuable when you’re taking standardized tests. If you have time, draw a diagram. I found this one on the first google result (I'm on my phone so it’s not easy to draw one of my own).

In this question, the right turn and the turn that takes the bus back to the main road aren't important. You only need to consider the distances that are driven on the main road or parallel to the main road. These distances are the 25km, 25km, and 35km in the diagram. So if you subtract them from 150, you find the answer of 65km.

Ah, a summation problem. We’re learning this right now, so this is perfect timing.

So, a ball goes down and up 36 meters, then up 24, down 24, up 16, down 16, and so on for infinity.

Basically it looks like this if you’re more visual.

So you initially drop it from 36 feet, then 24, then 16, and so on infinitely. Basically, you're doing 36 + 24 + 16 + …

There is a formula for when you have an infinite amount of numbers to sum up, which is [math]S_∞ = a_1/(1-r)[/math] where r is the rate, or 2/3.

So you take the first term excluding 36, and divide that by 1/3, which gives you 72. However, you have to add back the 36, and when you do, you get 108.

Here are a few resources to help you out.

http://www.classzone.com/eservices/home/pdf/student/LA211DBD.pdf (example 4)

SOLUTION: A reboncing ball rebounces each time to a height equal to one half the height of the previous bounce. If it is dropped from a height of 16 mtrs. find the total distance it has trav

They aren’t exactly the same problem but they’re close enough to help you understand it, because I’m horrible at explaining.

This is usually written as [math]O(g)[/math] (without the bar through the O), and it is easier to write so I will use that notation.

a) Reflexivity: [math]fOf[/math] follows from the definition for any value of [math]x_0[/math] and [math]M=1[/math]
Transitivity: Assume [math]fOg[/math] and [math]gOh[/math]. Then, define [math]x_1,M_1,x_2,M_2[/math] so that [math]\forall x \ge x_1 : |f(x)| \le M_1 |g(x)|[/math] and [math]\forall x\ge x_2:|g(x)|\le M_2 |h(x)|[/math]. Then, for [math]x_3=\max\{x_1,x_2\}[/math] and [math]M_3=|M_1M_2|[/math] we have that [math]x\ge x_3:|f(x)|\le M_1 |g(x)|\le M_3|h(x)|[/math].
For [math]O[/math] to be a (partial) order, we must have that [math]fOg[/math] and [math]gOf[/math] imply [math]g=f[/math]. A counter-example is, for any function [math]f[/math] (not the zero function) and constant [math]a\ne 1[/math], then [math]fOaf[/math] (by any value of [math]x_0[/math] and [math]M=a[/math]) and [math]afOf[/math] (with [math]M=1/a[/math]) but [math]f\ne af[/math].
As a bonus, for [math]O[/math] to be a total order, all functions must be comparable. A counter-example is that neither [math]sinO0[/math] nor [math]0Osin[/math] (where [math]0[/math] denotes the zero function, i.e. the constant function that is 0 for any value of the argument)

b) (see [1]) Two parts: => and <=
part 1: [math]n \le m \Rightarrow x^nOx^m[/math]
Take [math]x_0=1[/math] and [math]M=1[/math]. Then we need to prove that [math]\forall x>1:x^n\le x^m[/math]. Divide both sides by [math]x^n[/math] and note that [math]1\le x^(m-n)[/math] follows from the fact that [math]x>1[/math] and [math]n\le m[/math].
part 2: [math]x^nOx^m \Rightarrow n \le m[/math]
It is given that, for some [math]x_0,M[/math], we have [math]\forall x\ge x_0:|x^n|\le M |x^m|[/math]. From this, it also follows that the same statement is true if we replace [math]x_0[/math] by [math]x_1=\max\{x_0,1\}[/math] and [math]M[/math] by [math]M_1=\max\{M,1\}[/math], i.e. [math]\forall x\ge x_1:|x^n|\le M_1 |x^m|[/math]. Both sides of the equation are positive, so we can remove the abs. Dividing both sides by [math]x^n[/math], we have that [math]1\le x^(m-n)[/math]. Since [math]x\le 1[/math], this is true iff [math]n\le m[/math].

c) First, introduce a little trick: prove that if [math]fOg[/math], also [math](f+g)Og[/math]. Define [math]x_1,M_1[/math] as the [math]x_0,M[/math] associated with [math]fOg[/math]. Then [math]\forall x\ge x_o:|f(x)+g(x)|\le |f(x)|+|g(x)|[/math][math] \le M_1 |g(x)| = (M_1+1)|g(x)|[/math]. Thus, [math](f+g)Og[/math]. An analogous proof will also show that [math]gO(f+g)[/math].
The rest of the proof is best made by induction. Denote by [math]n[/math] the degree of [math]p[/math] (for general [math]p[/math]). For [math]n=0[/math], both [math]p[/math] and [math]x^n=x^0[/math] are constants, so both [math]pOp^n[/math] and [math]p^nOp[/math] are trivially true (see [4]).
For the induction step, denote by [math]p_*[/math] the polynomial that has the same terms as [math]p[/math] except the highest degree term. Then according to the induction hypothesis, [math]p_*Ox^{n-1}[/math] and [math]x^{n-1}Op_*[/math]. Now denote by [math]p^*[/math] the one-term polynomial consisting of only the highest-degree term, i.e. [math]p^*=p-p_*[/math]. Then, we know that [math]p^*Ox^n[/math] and [math]x^nOp^*[/math] (see [4]). Finally, from the proof made in the beginning of c), and the fact that [math]p_*Ox^{n-1}Op^*[/math], we may add [math](p^*+p_*)Ox^n[/math], and also [math]x^nO(p^*+p_*)[/math], which is equivalent to what we were trying to prove.

Bonus: Denote [math]n\equiv \deg{p_1}[/math], [math]m\equiv\deg{p_2}[/math]
=>: given that [math]p_1Op_2[/math], we also have [math]x^nOp_1Op_2Ox^m[/math]. Thus, from b), [math]n\le m[/math]
<= given that [math]n\le m[/math], we know that [math]x^nO x^m[/math] from b). Then, [math]p_1Ox^nOx^mOp_2[/math].

[1] There is a slight mistake in the question: it is not given that [math]m,n\ge 0[/math]. For negative exponents, the described function (see [2]) is not defined on [math]\mathbb{R}[/math] (the domain misses zero). Technically, the operator [math]O[/math] was only defined for functions with domain and codomain [math]\mathbb{R}[/math]. I will assume that the exponents may be of any real value and a natural extension of the definition of O.
[2] To be completely nitpicky, [math]x^m[/math] is an expression, not a function, but from context we can assume that the author meant the function described by the expression. The notation is clear and I will also adopt it, instead of the cumbersome but technically correct [math](x\mapsto x^m)[/math].
[3] [math]\circ[/math] denotes function composition, i.e. [math]\exp\circ f[/math] is the function so that [math](\exp\circ f)(x)=exp(f(x))[/math] for all [math]x[/math].
[4] In fact, we have already discussed a more general case in a), namely that [math]fOaf[/math] and [math]afOf[/math].

Jane has [math]\approx 3.14159[/math] potatoes left.

Proof:
Starting from the beginning, we see that John takes half of the apples from the available 97 grapes of the watermelon tree. Alex becomes adamant at this point and demands half of the almonds John has. John is equally adamant and finally settles in giving 1/3 of his almonds as Alex blackmails him of revealing John's affair with Jane.

Now comes the reason why Jane has 3.14159 potatoes. Jane was never a fan of oranges. It reminded her of the bad relationships she had with Alex and John, who were better of loving fruits than women. This is seen from their obsession in possessing fruits, especially apples and almonds.

Jane, was a huge fan of potatoes. She had infinite potatoes in many of her barns. One day, she decided to create a circle. She laid the potatoes in a circle. All of a sudden, infinitesimal human beings approached her and demanded some potatoes, since she had many. She decided to give some. She asked them to stand on the diameter of the circle she made from the potatoes and divided them equally among these humans.

Let's do a quick calculation. She gave the potatoes in the circle to the people standing on its diameter. this gives (Circumference of potato circle / Diameter) = 3.14159 potatoes.

Q.E.D.

Draw it out on a grid - or follow the logic

Assume the first bus starts at (0,0), and the 2nd bus starts at (150,0) - I assume that the main road runs horizontally along the y = 0 axis.

‘The 1st bus goes for 25km and then turns right and runs for 15km’ - so it reaches : (25,-15). ‘it then turns left and goes 25km’ : that takes it to (50,-15).

‘It then takes the direction back to reach the main road.’; the meaning here isn’t clear but I will assume it means goes directly back to the main road - so this will read the road by turning left and going ‘up’ to (50,0).

‘Meanwhile the 2nd bus has runs 35km along the main road’- so this bus is now at (115,0).

So what is the distance between (50,0) and (115,0) ? It is 65 km …

No, both answers are off slightly.

I will assume that arrowed segments are noted as being parallel to each other.
I note that no indication is made whether Y forms right angles, and assume not.

If arrowed-segments are parallel, the 50[math]^\circ[/math] angle implies that the 50 km segment divides the 150[math]^\circ[/math] angle into 50-degree and 100-degree subangles. Because a straight segment has 180 degrees, this we can infer that the triangle’s interior angle is 80[math]^\circ[/math].

We can apply the a similar process to the N[math]^\circ[/math] issue. But we will come back to that.

At this point, we know 2 sides and an angle. We can use the law of cosines to determine Y. That is [math]Y = \sqrt{ 50^2 + 40^2 - 2*40*50*cos(80) } = 58.36 km[/math]

From here, we can use the law of sines to solve the triangle. Work omitted:

Right-most interior angle: 57.54[math]^\circ[/math]

This doesn’t quite tell us what N is. Using that 40 km segment cross both parallel lines, we can see that the top-most exterior angle on the right arrow segment must be 180–150 = 30[math]^\circ[/math]

From this we can discern that N = 360 - 30 - 57 .54 = 272.4[math]^\circ[/math]

First, make both equations equal each other which makes 2^(x + 1) + 3 = 2^(2x) - 5.

Next, subtract 3 from both sides to get 2^(x + 1) = 2^(2x) - 8.

Then, represent -8 as a power of two which makes 2^(x + 1) = 2^(2x) - 2^3.

Then, take away all of the bases which results in x + 1 = 2x - 3.

Next, subtract 2x from both sides which gives you -x + 1 = -3.

Then, subtract 1 from both sides which makes -x = -4.

Next, multiply both sides by -1 to get x = 4.

Then, plug 4 in for x in 2^(x + 1) + 3 to get 2^(4 + 1) + 3.

Next, add 4 and 1 to get 5 which make the equation 2^5 + 3.

Then, exponentiate 2 by 5 to get 32 which means the equation is now 32 + 3.

Next, add 32 and 3 to get 35 which means f(x) is 35.

This means that the answer is (4,35).

This depends on what you mean by no answer. Some questions may not have an answer in the sense that an answer cannot possibly exist. simple examples would be inconsistent systems of linear equations, e.g. for what values of x and y do y = x and y = x + 1 intersect.

Other questions with no answers might just be conjectures that no mathematician has as of yet proven or disproven, like the previously noted question of P = NP, or other millenium prizes. Some such questions, though not technically proven, are fairly well understood. Fermat’s Last Theorem went famously unproven for almost 400 years, although by that time the betting odds were probably in favor of its truth.

Conversely, some theorems may be provable but not fully understood, for example the first proof of the 4 color map theorem was criticized for not providing much meaningful and intuitive insight/understanding of the problem, and being hard to verify due to its cumbersome length.

To further complicate things, some questions can be proven to be impossible to prove or disprove, like Zorn’s Lemma (a seemingly intuitive statement) under traditional axioms. In these cases, new axioms (read: accepted unprovable truths) may need to be adopted to prove one way or the other. In the case of Zorn’s Lemma, most mathematicians adopt an equivalent statement (the Axiom of Choice, “equivalent” in the sense that Zorn’s Lemma can prove AoC or vice versa), though this axiom is in a certain sense controversial because it gives rise to certain counter-intuitive conclusions: most famously the banach-tarski paradox. However, the axiom is generally accepted regardless.

For more info on the subject of solvable/unsolvable questions, read about axioms, logic, or set theory.

Not quite, assuming my calculator is correct.

The first step should be to find as many angles as you can using the rules of parallel lines. You should be able to find the remaining two angles around the top vertex, and the angle between the 40 km line and the vertical line (due North?) by this method.

Now you have enough information to find Y, using the cosine rule:

[math]Y^2=(40 \mathrm{km})^2 + (50 \mathrm{km})^2 - 2 \times 40 \mathrm{km} \times 50 \mathrm{km} \times \cos{y}[/math]

where y is the angle inside the top vertex of the triangle, opposite the side Y.

Now you can use Y, y and the remaining sides to find the other angles inside the triangle, using the sine rule.

Once you’ve done this, you should be able to find N using the fact that angles around a point must add to 360 °.

If you want to check your work, I get N = 272.456 ° (3 dp).

Question #7:

[math]f (x)[/math] is a function with a domain of [math]\{x \ge -4, x \in \R\}[/math] and a range of [math]\{y < 2, y \in R\}[/math]. Suppose a new function, [math]g(x)[/math] is obtained from [math]f (x)[/math] by performing the following transformations:

• A reflection in (across) the y-axis
• A vertical compression factor of 1/3 and horizontal stretch by a factor of 7
• Shift the graph 2 units down and 1 unit to the right

1. Give an equation for [math]g(x)[/math] in terms of [math]f (x)[/math]
2. Find the domain and range of [math]g(x)[/math]

Determine Function g(x)

Modify the domain and range by each step in the transformations list.

Example: reflection in (across) the y-axis

The domain of [math]f(x), \{x \ge -4, x \in \R\}[/math] shows the points parallel to the x-axis. Flip them backward to get a reflection in (across) the y-axis.

This makes the domain of [math]g(x), \{x \le +4, x \in \R\}[/math] when it is reflected across the y-axis.

The rest is left to the reader.