Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?

• May 27th 2011, 04:05 PM
stacey
Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?
Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?

So I'm able to do the first two steps
f(g(sinx))
f(1/sin)

But I'm not sure what the last step does. Normally something enclosed inside ||, such as |x| means the magnitude of x, right? So what does that operator mean in the context of this problem? How does y=1/sinx differ from y=|1/sinx|? Obviously, this can be written as y=|csc|.

b) Describe what each composition does in the order that it is applied. Again, I'm fine until the last function, where I don't know what the operator does.

• May 27th 2011, 04:38 PM
Also sprach Zarathustra
Quote:

Originally Posted by stacey
Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?

So I'm able to do the first two steps
f(g(sinx))
f(1/sin)

But I'm not sure what the last step does. Normally something enclosed inside ||, such as |x| means the magnitude of x, right? So what does that operator mean in the context of this problem? How does y=1/sinx differ from y=|1/sinx|? Obviously, this can be written as y=|csc|.

b) Describe what each composition does in the order that it is applied. Again, I'm fine until the last function, where I don't know what the operator does.

h(g(f(x)))=h(g(sin(x)))=h(1/sin(x))=|1/sin(x)|=1/|sin(x)|=|cosec(x)|

There is a difference between cosec(x) and |cosex(x)|.

Start by drawing y=cosec(x) and try to understand what the operation | | doing to cosec(x).

(Maybe you should draw the function f(x)=x tnd afterwards to draw g(x)=|x|, it will give you a clue)
• May 29th 2011, 12:13 PM
rtplol
Quote:

Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?
Let's start with the inside, and work our way out. g(f(x)) is the same as g(sin(x)), which is the same as 1/sin(x). Our final result is then h(1/sin(x)). When you subsitute the x in |x| with 1/sin(x), what do you get?
• May 29th 2011, 02:19 PM
Soroban
Hello, stacey!

Quote:

$\text{Given: }\:f(x)\,=\,\sin(x),\;\;g(x)\,=\,\tfrac{1}{x},\;\; h(x) \,=\, |x|$

$\text{Find: }\:h(g(f(x)))$

But I'm not sure what the last step does.
Normally something enclosed inside ||, such as |x| means the magnitude of x, right?
So what does that operator mean in the context of this problem?
How does $y\,=\,\tfrac{1}{\sin x}$ differ from $y\,=\,\left|\tfrac{1}{\sin x}\right|$?
Obviously, this can be written as $y \,=\,|\csc x|$ . Yes!

b) Describe what each composition does in the order that it is applied.
Again, I'm fine until the last function, where I don't know what the operator does.

Evidently, you're not familiar with "absolute values".
(I'm not sure how that happened to you.)

$|x|$ means: take the positive value of $x.$

. . $\begin{array}{ccc}\vdots && \vdots \\ |\text{-}2| &=& 2 \\ |\text{-}1| &=& 1 \\ |0| &=& 0 \\ |1| &=& 1 \\ |2| &=& 2 \\ \vdots && \vdots\end{array}$

When graphing, any portion below the x-axis is reflected upward above the x-axis.

Quote:

c) sketch the curve.
My curve looks like a csc function. .Is this correct?

Not quite . . . It is the cosecant graph, but everything is "made positive".

It looks something like this:

. . $\begin{array}{c} | \\ \cup \cup | \cup \cup \\ | \\ --+-- \\ | \end{array}$