Question

Suppose that u(t) = 1/t^2+1. Find simpler functions f(t) and g(t) such that u(t) = f(g(t))

Answer 1

There are multiple answers to this sort of question. The most obvious one would be
\[\begin{cases}f(t)=\dfrac{1}{t^2+1}\\
g(t)=t\end{cases}\]
which works because
\[u_1(t)=f(g(t))=f(t)=\frac{1}{t^2+1}\]
but something tells me you're expected to use something different. Perhpahs
\[\begin{cases}f(t)=\dfrac{1}{t}\\
g(t)=t^2+1\end{cases}\]
so that
\[u_2(t)=f(g(t))=f(t^2+1)=\frac{1}{t^2+1}\]
Or you can use
\[\begin{cases}f(t)=\dfrac{1}{t+1}\\
g(t)=t^2\end{cases}\]
so
\[u_3(t)=f(g(t))=f(t^2)=\frac{1}{t^2+1}\]
The point is thata there are many ways to do this.

Answer 2

wow, there is more than one answer to this

Answer 3

Yes, you can even make the component \(f\) and \(g\) as complicated as you want, so long as the composition results in the same function.

Answer 4

so what I have to do is come up with f(t) and g(t) that is going to give f(g(t))

Answer 5

Yep

Answer 6

what if I had f(t) as 1 and g(t) as t^2+1?

Answer 7

No, that wouldn't work because \(f\) is a constant function. If \(f(t)=1\), it doesn't matter what you use for \(g\) because
\[f(g(t))=f(t^2+1)=1\not=\frac{1}{t^2+1}\]

Answer 8

Think of it this way: if there's no \(t\) in the functional expression for \(f(t)\), then it won't work. The variable must be included.

Answer 9

so it would have to be f(t) = 1/t and g(t) = t^2+1

Answer 10

Yes

Answer 11

what is u(t) then?

Answer 12

that looks to me like a typo

Answer 13

\(u(t)\) will always be \(\dfrac{1}{t^2+1}\)

Answer 14

It's a more compact form of \(f(g(t))\)

Answer 15

the first answer you gave to me would be simpler... but I think we have to use the second answer

Answer 16

It's your choice. Like I said, many possible answers to this one.

Answer 17

And choosing the "simplest" one doesn't really make sense. How simple it is would depend on the context.

Answer 18

yes, you are right... my exam on this material is coming up tuesday and I don't feel ready

Answer 19

Well if my opinion means anything, I suppose I'd call my \(u_2(t)\) the simplest choice. Good luck.

Answer 20

ok I was looking over the u2t solution... f(t^2+1) is the same as 1(t^2+1)?

Answer 21

1/(t² + 1), yes

Answer 22

Since \(f(t)=\dfrac{1}{t}\), and \(g(t)=t^2+1\), then \(f(g(t))=f(t^2+1)=\dfrac{1}{t^2+1}\). Basically, you replace any \(t\) you see with \(t^2+1\).

Answer 23

ok I understand that part

Answer 24

You might benefit from being exposed to other examples.

Say \(f(t)=t^2\) and also that \(g(t)=t^2\). Then,
\[f(g(t))=f(t^2)=\left(t^2\right)^2=t^4\]
Makes sense?

Answer 25

yes

Answer 26

so you find out f(t) and g(t) if they are not given and then you replace t with the f(t)

Answer 27

Be careful with how you phrase that. If you want to find \(f\) and \(g\) such that \(f(g(t))\) matches up with the desired function, then you replace \(t\) in \(f\) with \(g(t)\).

However, if you want to match up \(g(f(t))\) with the given function, then you replace \(t\) in \(g\) with \(f(t)\).

Answer 28

so if I have f(t) = t^2+2t and g(s) = 2s+1
f(g(s)) = f(t^2+2t)=(2s+1)=t^2+2s+1?

Answer 29

I am probably off...

Answer 30

The variable of the resultant function should always be the one given in the "innermost" function; in this case, \(s\).

So,
\[f(t)=t^2+2t~~~~~~~~g(s)=2s+1\\
f(g(s))=f(2s+1)=(2s+1)^2+2(2s+1)=4s^2+8s+3\]

Answer 31

I am taking notes. You have a good way of explaining things.

Answer 32

It is often difficult to get concise notes out of a 45 minute class period.