# TrigonometryComposition of functions!

#### Alaba27

##### New member
This question is killing me. I'm finding it difficult to do and it's a problem with my homework.

Given that $$\displaystyle f(x)=2x^2-x+1,\,g(x)=2\sin(x)\text{ and }h(x)=3^x$$, determine the following. You need not simplify the expressions.

$$\displaystyle f(g(-\pi))=?$$

$$\displaystyle \left(h^{-1}\circ f \right)(x)=?$$

$$\displaystyle g(f(h(x)))=?$$  Last edited by a moderator:

#### MarkFL

Staff member
Let's begin with the first one...what is $$\displaystyle g(-\pi)$$?

#### Alaba27

##### New member
Let's begin with the first one...what is $$\displaystyle g(-\pi)$$?
I thought it was (0,1), but I'm not sure.

#### MarkFL

Staff member
You are only interested in the value the function returns not a point in the plane.

$$\displaystyle g(-\pi)=2\sin(-\pi)=-2\sin(\pi)$$

What is $$\displaystyle \sin(\pi)$$ ?

#### Alaba27

##### New member
You are only interested in the value the function returns not a point in the plane.

$$\displaystyle g(-\pi)=2\sin(-\pi)=-2\sin(\pi)$$

What is $$\displaystyle \sin(\pi)$$ ?
That's 0.

#### MarkFL

Staff member
Yes! So now, what is $$\displaystyle f(0)$$ ?

#### Alaba27

##### New member
Yes! So now, what is $$\displaystyle f(0)$$ ?
1! So it is (0,1)?

#### MarkFL

Staff member
It is just 1, when you write (0,1) this notation means an ordered pair, usually representing a point in a plane. I would write:

$$\displaystyle f(g(-\pi))=f(0)=1$$

Now for the second. We need to find $$\displaystyle h^{-1}(x)$$. Do you know how to find the inverse of a function and how to check your work to make sure you did it correctly?

#### Alaba27

##### New member
It is just 1, when you write (0,1) this notation means an ordered pair, usually representing a point in a plane. I would write:

$$\displaystyle f(g(-\pi))=f(0)=1$$

Now for the second. We need to find $$\displaystyle h^{-1}(x)$$. Do you know how to find the inverse of a function and how to check your work to make sure you did it correctly?
Alright. But I don't really understand how to do the second question. The thing is that I have to be done this question within the next 20 minutes because my tutor is only going to be available for a little bit today. #### MarkFL

Staff member
Well, we best get busy then...and I can't simply give you the answers because you have an impending deadline. My best advice is to ask for help earlier. I will be happy to stand in for your tutor to help you get these done, but I want to make sure you understand how to do them for yourself. Our goal is to make sure people gain a better understanding.

Do you know how to find the inverse of a function?

#### Alaba27

##### New member
Well, we best get busy then...and I can't simply give you the answers because you have an impending deadline. My best advice is to ask for help earlier. I will be happy to stand in for your tutor to help you get these done, but I want to make sure you understand how to do them for yourself. Our goal is to make sure people gain a better understanding.

Do you know how to find the inverse of a function?
I know, and I'd rather understand how to do the work instead of just getting answers! I know how to find the inverses of functions, but I've never done it in a composition function.

#### MarkFL

Staff member
We need not worry about the composition yet, all we need first is to find the definition of $$\displaystyle h^{-1}(x)$$. Can you find this?

Once we have it, then we will proceed to find the given composition.

#### Alaba27

##### New member
We need not worry about the composition yet, all we need first is to find the definition of $$\displaystyle h^{-1}(x)$$. Can you find this?

Once we have it, then we will proceed to find the given composition.
It's cube-root x.

#### MarkFL

Staff member
That would be correct if $$\displaystyle h(x)=x^3$$, but we have $$\displaystyle h(x)=3^x$$. You are going to need to convert from exponential to logarithmic form.

#### Fermat

##### Active member
This question is killing me. I'm finding it difficult to do and it's a problem with my homework.

Given that $$\displaystyle f(x)=2x^2-x+1,\,g(x)=2\sin(x)\text{ and }h(x)=3^x$$, determine the following. You need not simplify the expressions.

$$\displaystyle f(g(-\pi))=?$$

$$\displaystyle \left(h^{-1}\circ f \right)(x)=?$$

$$\displaystyle g(f(h(x)))=?$$

View attachment 846
View attachment 847

$g(-\pi)=0$ so $f(g(-\pi))=1$. $h^{-1}(x)=log_{3}(x)$ so $h^{-1}(f(x))=log_{3}(2x^2-x+1)$. Finall, $f(h(x))=2.9^x-3^x+1$ so $g(f(h(x)))=2sin(2.9^x-3^x+1)$