Given b(x)=|x+4|, what is b(-10)?
A.-10
B. -6
C. 6
D. 14

- anonymous

Given b(x)=|x+4|, what is b(-10)?
A.-10
B. -6
C. 6
D. 14

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- chestercat

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- SolomonZelman

at first, plug in -10 for x, into the \(|x+4|\).

- anonymous

-6?

- SolomonZelman

no, hold on, when you plug in -10 for x, you get:
\(\small b(x)=\left|x+4\right|\)
\(\small b(\color{red}{-10})=\left|\color{red}{-10}+4\right|\)
\(\small b(\color{red}{-10})=\left|-6\right|\)

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## More answers

- SolomonZelman

What does "absolute value of a", written as "|a|" mean?
Roughly speaking, you can define |a| as a distance from 0 to a
(how much do you have to walk from a to 0)

- anonymous

so would it be 6 instead of negative 6?

- SolomonZelman

|dw:1437934269242:dw|

- SolomonZelman

So, yes the answer is 6.

- SolomonZelman

in other words, the distance from 0 to -6 is 6.
why is it so that 0-6=-6 (not just 6)
because the minus by 6 indicates the direction (that it is going to the left - if you look at the number line),
but absolute value of -6, or |-6| is 6

- SolomonZelman

So, lets review the work we have to do:

- SolomonZelman

\(\small b(x)=\left|x+4\right|\)
\(\small b(\color{red}{-10})=\left|\color{red}{-10}+4\right|\)
\(\small b(-10)=\left|-6\right|\)
\(\small b(-10)=-6\)

- SolomonZelman

if you want to ask anything you are always welcome to do so...:)

- anonymous

That helps a lot! Thank you so much! Would you be able to help with another problem?

- anonymous

@SolomonZelman

- SolomonZelman

Yes, perhaps, but I am helping another student currently.
I personally don't care tho', if you post it in this thread or another one, so you can keep this post.

- anonymous

For which pair of functions is (f Â° g)(x) =x?
A) f(x) =x^2 and g(x) = 1/x
B) f(x)=2/x and g(x) = 2/x
C) f(x)= x-2/3 and g(x)= 2-3x
D) f(x)=1/2x-2 and g(x)= 1/2x+2

- SolomonZelman

I will show you an example on two different functions.

- SolomonZelman

So, lets say my function f(x) and g(x) are the following:
\(\large\color{blue}{ \displaystyle f(x)=3x^3+2 \\[0.5em]}\)
\(\large\color{red}{ \displaystyle g(x)=\frac{5}{x} }\)
--------------------------------------------
A notation of: \((\color{blue}{f}^\circ \color{red}{g})(x)\)
means the same as \(\color{blue}{f(~\color{red}{g(x)}~)}\)
in other words, you take a function g(x), and plug it instead of x, into the f(x).
--------------------------------------------
How would it actually go overhere?
\(\large\color{black}{ \displaystyle (\color{blue}{f}^\circ \color{red}{g})(x)=\color{blue}{f(~\color{red}{g(x)}~)}=\color{blue}{3\left(\color{red}{\frac{5}{x}}\right)^3+2} }\)
see how I am plugging the g(x) instead of x, into f(x)?
now, we will simplify that:
\(\large\color{black}{ \displaystyle (\color{blue}{f}^\circ \color{red}{g})(x)=\color{blue}{f(~\color{red}{g(x)}~)}=3\left(\frac{5}{x}\right)^3+2 }\)
\(\large\color{black}{ \displaystyle (\color{blue}{f}^\circ \color{red}{g})(x)=\color{blue}{f(~\color{red}{g(x)}~)}=3\frac{5^3}{x^3}+2 }\)
\(\large\color{black}{ \displaystyle (\color{blue}{f}^\circ \color{red}{g})(x)=\color{blue}{f(~\color{red}{g(x)}~)}=3\frac{125}{x^3}+2 }\)
\(\large\color{black}{ \displaystyle (\color{blue}{f}^\circ \color{red}{g})(x)=\color{blue}{f(~\color{red}{g(x)}~)}=\frac{375}{x^3}+2 }\)

- SolomonZelman

so all you are doing for (fÂº g)(x), is that you plug in the function g(x) instead of x --> into the f(x).

- SolomonZelman

my function is more rather complicated, but all you need to do is to find this (fÂº g)(x),
and which ever of the options will give you a result/answer of x, that is your answer.

- SolomonZelman

I will respot the question:
----------------------
For which pair of functions is (f Â° g)(x) =x?
A) f(x) =x^2 and g(x) = 1/x
B) f(x)=2/x and g(x) = 2/x
C) f(x)= x-2/3 and g(x)= 2-3x
D) f(x)=1/2x-2 and g(x)= 1/2x+2

- SolomonZelman

now, lets find (fÂº g)(x) for each answer choice together:
which one do you want to start from?

- anonymous

Which do you think would work better?

- anonymous

We can just start from A

- SolomonZelman

yes, lets do that.

- SolomonZelman

f(x) =xÂ²
g(x) = 1/x
(fÂº g)(x) = f(g(x)) = (1/x)Â²=1Â²/xÂ²=1/xÂ²

- SolomonZelman

so is A the answer or not?

- anonymous

no!

- SolomonZelman

yes, it is not the answer.

- SolomonZelman

lets do the next one, but this one i will ask you to take a shot on

- SolomonZelman

f(x)=2/x
g(x) = 2/x
(fÂº g)(x) = ?
(go ahead)

- anonymous

(2/x)^2 =x^2=2/x^2
is that how we set it up

- SolomonZelman

(fÂº g)(x) is same as f(g(x))
so you just plug in the g(x) (in this case, 2/x), instead of x, into the f(x).
so
f(x) = 2/x
(fÂº g)(x) = f(g(x))=2/(2/x)

- SolomonZelman

all we did is that we took the x in f(x), and replaced it by the function g(x).
(i.e. replaced the x in the function f(x), but 2/x)

- SolomonZelman

now, you need to simplify this. can you do that?

- anonymous

I got 1/x is that correct?

- SolomonZelman

|dw:1437937373887:dw|

- SolomonZelman

|dw:1437937442588:dw|

- anonymous

then would it just be x? plain and simple?

- SolomonZelman

yes

- SolomonZelman

it is =x

- SolomonZelman

and thus B is the answer

- anonymous

AHA! Thank you so much

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