VCE Methods Question Thread!

[kinslayer]

Anyone?

The range of h depends on what f + g looks like. For example, if f(x) = -g(x) then the range of h(x) is {0}, no matter what the ranges of f or g are.

The best thing is to just draw the graph.

[I_I]

I have a question on exponential

Q5. Given f(x)=e^x-a+1 and a is R⁺, find the value of a such that the graphs of y=f^-1(x) and y=f(x) touch each other

This btw is a tech free question

Q6. Given f(x)= -(x+1)²(x-5)²(x-2)³ and g(x)= -f(x-a), g(-x)=-g(x)
The answer, btw is a=-2

Q3. let f(x)=2/(e^x-e^-x), x is R
FIND f^-1(x)

Q7. The question ke^2x -2e^x +k=0 has two solutions for x. Find the possible values of k.
I got the answer -1<k<1 but the answer is {k: -1<k<0} U {k: 0<k<1}

please help! Sorry for posting so many questions, but if anyone could please post the solutions (with working out please), I would be forever in their debt !!!!

[I_I]

for Q7, it's 'equation' not 'question' sorry

[lzxnl]

I have a question on exponential

Q5. Given f(x)=e^x-a+1 and a is R⁺, find the value of a such that the graphs of y=f^-1(x) and y=f(x) touch each other

This btw is a tech free question

Q6. Given f(x)= -(x+1)²(x-5)²(x-2)³ and g(x)= -f(x-a), g(-x)=-g(x)
The answer, btw is a=-2

Q3. let f(x)=2/(e^x-e^-x), x is R
FIND f^-1(x)

Q7. The question ke^2x -2e^x +k=0 has two solutions for x. Find the possible values of k.
I got the answer -1<k<1 but the answer is {k: -1<k<0} U {k: 0<k<1}

please help! Sorry for posting so many questions, but if anyone could please post the solutions (with working out please), I would be forever in their debt !!!!

Q5
Well you need the function to intersect its inverse first.
So exp(x-a) + 1 = x
I'd say the function needs to touch the line y = x as well because of the relationship between the gradients of a function and its inverse.
In other words, we also need to satisfy the equation f'(x) = exp(x-a) = 1, so x = a
In the first line, x = a means e(0) + 1 = a. a = 2
f(x) = exp(x-2) + 1

Inverse is g(x) = ln(x-1) + 2. g(2) = 2. Also g'(x) = 1/(x-1), g'(2) = 1 as needed.
Hence f(x) = g(x) at x = 2 and f'(x) = g'(x)

Q6
f(x) = -(x+1)^2 (x-5)^2 (x-2)^3? So somehow I'm meant to translate this function to get an odd function? Ok...
Well, the intercepts are at -1, 2 and 5. x = 2 is a stationary point of inflection. Instinct tells me that x = -1 is a local minimum and x = 5 is a local maximum, which is fine.
Seeing how the intercepts are spaced out so evenly, I'm going to introduce a translation just so that the middle stationary point is at the origin.
Let u = x-2, so x = u + 2
f(u) = -(u+3)^2 (u-3)^2 u^3. This is an odd function because replacing u with -u does not affect the two perfect square brackets, but it flips the sign of u^3. Hence replacing x with u + 2 gives an odd function.

So likewise, replacing x with x + 2 gives an odd function => x + 2 = x - a, a = -2

Q3
f(x) = 2/(e^x - e^-x)
Let y = f(x). Swap x and y for inverse and solve for y
x = 2/(e^y - e^-y)
e^y - e^-y = 2/x
e^2y - 2e^y/x - 1 = 0
(e^y - 1/x)^2 - 1 = 1/x^2
(e^y - 1/x) = +-sqrt(1 + 1/x^2)
e^y = 1/x +- sqrt(1 + 1/x^2)

sqrt(1+1/x^2) > 1/x for all x so take the positive square root (e^y > 0)
So y = ln(1/x + sqrt(1+1/x^2))

Q7
Only problem with k=0 is that your equation becomes -2e^x = 0 which has no solution. Hence you must exclude k=0 from -1<k<1.

[I_I]

Ahhh!!! Thank you so much! 

[chansena]

Hi!

Im stuck with this problem / question and I'm supper confusion (yes i said confusion )

for the fuction f(x) =X^2+2x and h(x)=3x+1 find:

F o h(x)

h o f(x)

f o h(3)

If you could be supper detailed that would be supper awesome !

Thanks !!

[warya]

Is there any actual use of composite functions? What's the purpose of it

And can someone explain why the domain of the first function must be the domain of the second? I mean, wouldn't the range of the first one become the range of the second one?

Thanks

Its the domain of the inside function that determines the domain of the composite function, this is because, if the numbers can't go into the inside function then it like wouldn't work as part of the function isn't defined for that domain. The range of the inside function must fit into the domain of the outside function as all of its outputs will become inputs of the outside function so everything just has to fit in to each other.

As for why the exist, idk

[keltingmeith]

Is there any actual use of composite functions? What's the purpose of it

And can someone explain why the domain of the first function must be the domain of the second? I mean, wouldn't the range of the first one become the range of the second one?

Thanks

The reason they exist is because we compose functions a lot, so we need to have some strict rules and guidelines for them to work. For example, the standard deviation function (which you'll learn about in probabiliy) is a composition of the square root and variance functions:

[Ha_Nguyen]

Hey guys, how do I do this question?
Prove whether these statements are true, or find a counterexample to demonstrate that it is false

a) for any function f, f(xy) = f(x)+f(y)

thanks

[lzxnl]

Hey guys, how do I do this question?
Prove whether these statements are true, or find a counterexample to demonstrate that it is false

a) for any function f, f(xy) = f(x)+f(y)

thanks

f(x) = e^x is a counterexample
As are almost all polynomials with non-zero y intercept
Just show that, for instance f(xy) = e^(xy) which isn't the same as e^(x) + e^(y)
Easily verified by setting x = y = 0

[Apink!]

Q5
Well you need the function to intersect its inverse first.
So exp(x-a) + 1 = x
I'd say the function needs to touch the line y = x as well because of the relationship between the gradients of a function and its inverse.
In other words, we also need to satisfy the equation f'(x) = exp(x-a) = 1, so x = a
In the first line, x = a means e(0) + 1 = a. a = 2
f(x) = exp(x-2) + 1

Inverse is g(x) = ln(x-1) + 2. g(2) = 2. Also g'(x) = 1/(x-1), g'(2) = 1 as needed.
Hence f(x) = g(x) at x = 2 and f'(x) = g'(x)


Hi, for this question I still don't understand. Sorry for asking questions about someone else's question but I tried solving this too but I couldn't solve it as well and I don't get the explanation.
Could you please explain it again?

[Ha_Nguyen]

How do i write f(x) as a piecewise defined/hybrid function?

f(x)=|x-4|(x+2)
How do i show working for this?

[Zealous]

How do i write f(x) as a piecewise defined/hybrid function?

f(x)=|x-4|(x+2)
How do i show working for this?

Hint: When x<4, (x-4) is going to be multiplied by negative 1 as it is inside of the modulus function and (x-4) is negative over that domain. When x≥4, (x-4) is going to be the same as (x-4) is positive over that domain and the modulus operator has no effect on it. So you can split f(x) into two pieces, one piece when the (x-4) is multiplied by negative 1, and one piece where it isn't.

This graph may help illustrate it: https://www.desmos.com/calculator/jvjxrfxjbr

[lzxnl]

Q5
Well you need the function to intersect its inverse first.
So exp(x-a) + 1 = x
I'd say the function needs to touch the line y = x as well because of the relationship between the gradients of a function and its inverse.
In other words, we also need to satisfy the equation f'(x) = exp(x-a) = 1, so x = a
In the first line, x = a means e(0) + 1 = a. a = 2
f(x) = exp(x-2) + 1

Inverse is g(x) = ln(x-1) + 2. g(2) = 2. Also g'(x) = 1/(x-1), g'(2) = 1 as needed.
Hence f(x) = g(x) at x = 2 and f'(x) = g'(x)


Hi, for this question I still don't understand. Sorry for asking questions about someone else's question but I tried solving this too but I couldn't solve it as well and I don't get the explanation.
Could you please explain it again?

Consider y as a function of x, so y = f(x)
Let's assume it's differentiable. Moreover, let's assume it has an inverse g so that if x=f(y), then y=g(x)
Now by the chain rule, dy/du = dy/dx * dx/du
If we let u = y we get dy/dy = dy/dx * dx/dy
So dy/dx = 1/(dx/dy)

In the case of y = f(x), working out dx/dy is essentially working with dy/dx for the inverse function of f. Which means dx/dy for y=f(x) finds you the slope of the inverse function.
If a function touches its inverse, it must intersect its inverse AND have the same slope. As its slope must be the reciprocal of the slope of its inverse, the slope can either be 1 or -1 at the point of intersection. I just assumed that the slope was equal to 1.

^yes that's all intuition for me, hence why I didn't clarify that bit of working

[Ha_Nguyen]

find all solutions to |cos(2x)|=1/2 ?!?!
how does the mod affect this question