VCE Methods Question Thread!

[Tadd12345]

For an inverse function to exist, the function being inverted must be one-to-one. In the case of a truncus, it can have an inverse function if and only if it is restricted to a domain where it is one-to-one (i.e. for y = 1/(x^2), this domain may be x an element of (0, infinity) etc).

Thx! I was confused for a second because the answer to the multiple choice question in my textbook implied trunci do have inverses?! but then I noticed the
R+ - R restriction xD

[m2121]

Can someone explain how to solve this? Q2 a,b,c. I tried to solve question 2a by expanding what is given as shown from the question; both of those expressions first and then somehow to equate coefficients however, I couldn't get it.

[TheBigC]

Can someone explain how to solve this? Q2 a,b,c. I tried to solve question 2a by expanding what is given as shown from the question; both of those expressions first and then somehow to equate coefficients however, I couldn't get it.

Expanding and equating is the correct method: you may have incorrectly expanded, that is all:

expanding a*(x - 3)^3 + b gives a*(x^2 - 6x + 9)(x - 3) +b = a*(x^3 -3*x^2 -6*x^2 +18*x +9*x - 27) + b = a*x^3 - 9*a*x^2 + 27*a*x -27*a + b
Hence, a = 2 (as coefficient of x^3 is 2) and b = 5 as -27*2 +b = -49, thus b = -49 + 54 = 5.

Another, more haste method of solving this question, is simply to notice that 'a' must equal 2 as the coefficient of x^3 is 2 and by multiplying the right-hand side of the brackets being expanded and adding 'b' to yield the final coefficient of -49 as demonstrated:

a*(x - 3)^3 + b = a*(x - 3)*(x - 3)*(x - 3) + b,

a*-3*-3*-3 + b = -49, hence -27*a +b = -49, thus b = 5.

[TrueTears]

For the third part, if you could write it in the supposed form, then by expanding and equating coefficients, you will find that not all equalities can be satisfied (I leave the algebra as an exercise to you). In general, in the supposed form, you only have three parameters but a cubic depends on four parameters, so essentially you are missing a "degree of freedom".

[hawkrook]

I'm a bit confused about the process to find x-intercepts of sin and cos graphs when there are transformations. Let's consider the equation y = 2 sin 2(x - pi/3) - root 3 for x [0, 2pi]. I understand that to find the x-ints you need to adjust the domain by adding pi/3 to the min and max of the domain. But why do you do this but you don't adjust the domain by multiplying the domain due to the '2' in front of (x - pi/3)?

[TheBigC]

I'm a bit confused about the process to find x-intercepts of sin and cos graphs when there are transformations. Let's consider the equation y = 2 sin 2(x - pi/3) - root 3 for x [0, 2pi]. I understand that to find the x-ints you need to adjust the domain by adding pi/3 to the min and max of the domain. But why do you do this but you don't adjust the domain by multiplying the domain due to the '2' in front of (x - pi/3)?

Finding intercepts is quite simple, as long as you don't over-complicate the problem. All that must be recognised, albeit the presence of transformations, is that when y = 0, an x-intercept exists. For example, in your situation, let 0 = 2*sin(2*(x - pi/3)), thus,

0 = sin(2*(x - pi/3)).

sin(x) = 0 when x = 0, pi, 2*pi etc. hence,

2*(x - pi/3) = 0, pi, 2*pi, 3*pi. (* to see how I acquired these specific values (from 0 to 3*pi) look below)

Therefore, solving for x, we get:

x = pi/3, 5*pi/6, 11*pi/6 (for x between 0 and 2*pi).

(* how I acquired values between 0 and 3*pi)

You also asked about restricting the domain.

as x is an element of (0, 2*pi), then we know that we can only have x values between 0 and 2*pi, thence we must find the output values of 2*(x - pi/3) that are allowed:

2*(0 - pi/3) = -2*pi/3 and 2*(2*pi - pi/3) = 10*pi/3
Notice how my values previously were 0, pi, 2*pi and 3*pi, these lie within the restricted output of (-2*pi/3, 10*pi/3).

[mzhao]

I'm a bit confused about the process to find x-intercepts of sin and cos graphs when there are transformations. Let's consider the equation y = 2 sin 2(x - pi/3) - root 3 for x [0, 2pi]. I understand that to find the x-ints you need to adjust the domain by adding pi/3 to the min and max of the domain. But why do you do this but you don't adjust the domain by multiplying the domain due to the '2' in front of (x - pi/3)?

Hey hawkrook,

You most certainly do have to take into consideration the '2'. When solving your equation:
\begin{align*}
0 &= 2\sin(2(x-\frac{\pi}{3})) - \sqrt{3}, x \in [0, 2\pi]\\
\implies \frac{\sqrt{3}}{2} &= \sin(2(x-\frac{\pi}{3}))\\
\implies 2(x-\frac{\pi}{3}) &= \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}
\end{align*}
you are finding all the possible values that the expression inside sin() can be in order for the right hand side to equal √3/2. So to make sure you don't go outside of your domain for x, you "adjust the domain" in order to find the range of values that 2(x - π/3) must be between. Here's the way I like to do it:
\begin{align*}
x &\in [0, 2\pi]\\
\implies x-\frac{\pi}{3} &\in [-\frac{\pi}{3}, \frac{5\pi}{3}]\\
\implies 2(x-\frac{\pi}{3}) &\in [-\frac{2\pi}{3}, \frac{10\pi}{3}]
\end{align*}
Just like if you are manipulating equations, whatever you do on the left, you do on the right to both ends of your interval.

[userrrname]

Can someone please help me with this question. It's probably an "easy" question so the function is f(x)=1/27(2x-1)^3(6-3x)+1. The question is find the x coordinate of each of the stationary points of f and state the nature of each of these stationary points.

[darkz]

Can someone please help me with this question. It's probably an "easy" question so the function is f(x)=1/27(2x-1)^3(6-3x)+1. The question is find the x coordinate of each of the stationary points of f and state the nature of each of these stationary points.

Hey
So I'll just tell you the general idea of how to find the stationary points
- First, find the derivative of f(x)
- Let the derivative equal to 0 and solve for x
- Sub the solutions for x back into f(x) to find the y coordinate
- For finding the nature of the stationary point, you can either draw up a table and calculate the gradient of two x coordinates on either side of the stationary point by using f'(x). If its \_/ then its a local min, if its /-\ then its a local max, if its /-/ then its a stationary point of inflection. (/ means positive gradient, - means gradient = 0, \ means gradient is negative)
- The other method is to use the second derivative and simply sub in the solutions of x, if f''(x) < 0 then it is a local max, f''(x) > 0 then local min, if =0 then stationary point of inflection

[userrrname]

Hey
So I'll just tell you the general idea of how to find the stationary points
- First, find the derivative of f(x)
- Let the derivative equal to 0 and solve for x
- Sub the solutions for x back into f(x) to find the y coordinate
- For finding the nature of the stationary point, you can either draw up a table and calculate the gradient of two x coordinates on either side of the stationary point by using f'(x). If its \_/ then its a local min, if its /-\ then its a local max, if its /-/ then its a stationary point of inflection. (/ means positive gradient, - means gradient = 0, \ means gradient is negative)
- The other method is to use the second derivative and simply sub in the solutions of x, if f''(x) < 0 then it is a local max, f''(x) > 0 then local min, if =0 then stationary point of inflection

Haha thanks for the answer, but I know how to find the stationary points its just that my answer seems to be completely wrong. I just wanted to see someone else solve it to see what I'm doing wrong. Thanks for your help tho!

[Shadowxo]

Haha thanks for the answer, but I know how to find the stationary points its just that my answer seems to be completely wrong. I just wanted to see someone else solve it to see what I'm doing wrong. Thanks for your help tho!

Hi , Could you post your working? I can write up a solution but it would take a bit longer. If you would like a written solution, I just want to check that the equation is

[userrrname]

Hi , Could you post your working? I can write up a solution but it would take a bit longer. If you would like a written solution, I just want to check that the equation is

Hey, could you at least just do the differentiation part? If it wouldn't take too much of your time, because I feel like that's where I might of went wrong. Thanks so much in advance!!

Btw that is the correct equation

[RuiAce]

Hey, could you at least just do the differentiation part? If it wouldn't take too much of your time, because I feel like that's where I might of went wrong. Thanks so much in advance!!

Btw that is the correct equation

Check on WolframAlpha

[Lear]

Hi,
I seem to recall there was a quick trick on the cas to solve this involving graphing the binomial function and then splitting the screen by opening a table (Menu-7-1) and then scrolling down to see at what point it becomes 0.85. I can't seem to remember the exact steps, however.

Question is this 'The probability of winning a prize in a particular competition is 0.2. How
many tickets would someone need to buy in order to guarantee them a probability
of at least 0.85 of winning a prize?'

Am I recalling correctly? If so what were the steps again? I think I am entering my values into Binomial PDF incorrectly.

[jazzycab]

Hi,
I seem to recall there was a quick trick on the cas to solve this involving graphing the binomial function and then splitting the screen by opening a table (Menu-7-1) and then scrolling down to see at what point it becomes 0.85. I can't seem to remember the exact steps, however.

Question is this 'The probability of winning a prize in a particular competition is 0.2. How
many tickets would someone need to buy in order to guarantee them a probability
of at least 0.85 of winning a prize?'

Am I recalling correctly? If so what were the steps again? I think I am entering my values into Binomial PDF incorrectly.

Yes, you can certainly do this in the CAS, but remember when posting questions of this type on AN that there are different CAS calculators used across the state.
It seems, from your menu buttons, that you are using the TI-NSpire.
I think your problem is that you are using the binomPDF function, rather than the binomCDF function.
We have a binomial variable with \(n\) trials and a probability of success of \(p=0.2\), i.e. \(X\sim\text{Bi}\left(n,0.2\right)\) and we want \(\text{Pr}\left(X\ge1\right)\ge0.85\).
I have included screenshots below of how to set this up on the NSpire - otherwise, use the binomial expansion and solve:

This can be solved readily using logs or the CAS solving capabilities.