Year 12s: what will be the first way you research your future options? >>

Composite Functions

If f(x)=ax+b and f^(3)(x)=64x+21 find the values of a and b. Suggest a rule for f^(n)(x)

Would f^3(x) be a^3x+3b and f^n(x)= 4^n(x)+7n?

Posted from TSR Mobile

Reply 1

timebent

Original post by Mr Pussyfoot

If f(x)=ax+b and f^(3)(x)=64x+21 find the values of a and b. Suggest a rule for f^(n)(x)

Would f^3(x) be a^3x+3b and f^n(x)= 4^n(x)+7n?

Posted from TSR Mobile

It certainly fits so i believe it's correct

Reply 2

ghostwalker

Original post by Mr Pussyfoot

If f(x)=ax+b and f^(3)(x)=64x+21 find the values of a and b. Suggest a rule for f^(n)(x)

Would f^3(x) be a^3x+3b and f^n(x)= 4^n(x)+7n?

Posted from TSR Mobile

No.

f(x) = ax+b

So,

f^2(x) = f(f(x)) = a(ax+b)+b

and

f^3(x) = f(f(f(x))) = a(a(ax+b)+b)+b

And take it from there.

Reply 3

B_9710

It may be useful to prove your suggested rule by induction, quite easy.

Reply 4

Mr Pussyfoot

Original post by ghostwalker

No.

f(x) = ax+b

So,

f^2(x) = f(f(x)) = a(ax+b)+b

and

f^3(x) = f(f(f(x))) = a(a(ax+b)+b)+b

And take it from there.

Would the rule be a^(n)x+b(a^(n-1)+a^(n-2)...a^0)?

Posted from TSR Mobile

Reply 5

ghostwalker

Original post by Mr Pussyfoot

Would the rule be a^(n)x+b(a^(n-1)+a^(n-2)...a^0)?
Posted from TSR Mobile

Looks good. Notice that the constant term is imply "b" times a G.P., so can be simplified.

Reply 6

Mr Pussyfoot

Original post by ghostwalker

Looks good. Notice that the constant term is imply "b" times a G.P., so can be simplified.

I see, so the fully simplified version would come down to a^nx+b(1-a^n/1-a) I'm guessing

Posted from TSR Mobile

Reply 7

Mr Pussyfoot

And on another note, if a f(x)=1/(1-x), would f^2(x)= 1/(2+x)? Nevermind, this is incorrect lol

Posted from TSR Mobile

(edited 7 years ago)

Reply 8

Zacken

Original post by Mr Pussyfoot

And on another note, if a f(x)=1/(1-x), would f^2(x)= 1/(2+x)? Nevermind, this is incorrect lol

Posted from TSR Mobile

No, it wouldn't be. It would be

\displaystyle f(f(x) = \frac{1}{1 - f(x)} = \frac{1}{1 - \frac{1}{1-x}} = \frac{1-x}{1-x - 1} = \cdots

Reply 9

Mr Pussyfoot

ImageUploadedByStudent Room1469191557.462888.jpg

ImageUploadedByStudent Room1469192166.948401.jpg

I've gotten this far on question 15 but have no idea what to do next, should I substitute the values of x into f^-1(x)=f^2(x)?

Posted from TSR Mobile

Reply 10

Mr Pussyfoot

Original post by Zacken

No, it wouldn't be. It would be

\displaystyle f(f(x) = \frac{1}{1 - f(x)} = \frac{1}{1 - \frac{1}{1-x}} = \frac{1-x}{1-x - 1} = \cdots

Yeah I realised this after a while

Posted from TSR Mobile

Reply 11

Zacken

Original post by Mr Pussyfoot

I've gotten this far on question 15 but have no idea what to do next, should I substitute the values of x into f^-1(x)=f^2(x)?

Nope, nope, nope. This is why I always encourage writing

f(f(x)) = \frac{a}{f(x)} + b = \frac{a}{\frac{a}{x} + b} +b \neq \frac{a}{\frac{a}{x}} + b + b

Reply 12

ghostwalker

Original post by Mr Pussyfoot

I see, so the fully simplified version would come down to a^nx+b(1-a^n/1-a) I'm guessing

Posted from TSR Mobile

That's correct.

Though you can of course simplify further by putting in the actual values of a and b.

Then as a check set n=3, and see if if gives the desired result. You can also check n=1, to see it gives ax+b, for whatever your values of a,b are.

Reply 13

Mr Pussyfoot

Original post by Zacken

Nope, nope, nope. This is why I always encourage writing

f(f(x)) = \frac{a}{f(x)} + b = \frac{a}{\frac{a}{x} + b} +b \neq \frac{a}{\frac{a}{x}} + b + b

ImageUploadedByStudent Room1469224542.857465.jpg

Okay I arrived at this quadratic, should I follow on to solve for x? I notice that the coefficient of x^2 is similar to what the question is asking for, is there a link?

Posted from TSR Mobile

ImageUploadedByStudent Room1469224530.188341.jpg14.6KB

Reply 14

Zacken

Original post by Mr Pussyfoot

Okay I arrived at this quadratic, should I follow on to solve for x? I notice that the coefficient of x^2 is similar to what the question is asking for, is there a link?

Posted from TSR Mobile

You want the quadratic to be satisfied for all values of

x

, that is, you need all of the coefficients to be equal to 0 so that the whole thing is equal to 0, i.e: solving for

x

will give two values of

x

in terms of

a

and

b

such that the quadratic holds, you want the quadratic to be identically 0 for all

x

(think in terms of: you want it to be an identity not an equation), so anyway.

The quadratic is

\displaystyle (a+b^2)x^2 - b(a+b^2)x - a(a + b^2) = 0 \iff (a + b^2)(x^2 - bx - a) = 0

So you need all of

a+b^2 = 0

(edited 7 years ago)

Quick Reply

Latest

Articles for you

How to be ready for an apprenticeship interview

Who should you take to an open day?

University halls: 10 things you'll only learn after moving in

What is an LLM and how could it benefit your career?

Composite Functions

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you