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# Function Notation problem

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Given that f(0)=−7, f(1)=−9, f(2)=16 f(3)=13, f(4)=−6,

and g(0)=4, g(1)=3, g(2)=0, g(3)=2, g(4)=1,

evaluate the following:

(a) (f∘g)(0)=

(b) (f∘g)(1)=

(c) (f∘g)(2)=

(d) (f∘g)(3)=

(e) (f∘g)(4)=

Please help, I've been trying at it for hours and I still can't wrap my head around it, and if possible, please explain how you got to the answer? I really want to learn this well, as I feel that I'm doing one step incorrectly and therefore the whole thing is wrong

Feb 19, 2019
edited by Roxettna  Feb 19, 2019

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$$(f\circ g)(x) = f(g(x))\\ -\\ \begin{matrix} x &g(x) &f(g(x)) \\ 0 &4 &-6\\ 1 &3 &13 \\ \vdots \end{matrix}\\ \text{see how it works?}$$

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Feb 19, 2019
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Yes! thank you very much

Roxettna  Feb 19, 2019
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The notation is probably what is giving you trouble

( f ° g) (0)    means this   ⇒   f ( g(0) )

We first evaluate  g(0)   = 4

So...now we have

f ( g(0) ) =   f(4) = -6

Another way to see this is to work from right to left

So  put 0 into g  and we get  g(0) =  4

Then...put this into f  and we have f(4)   = -6

Does that make sense ???

Feb 19, 2019
#5
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Thank you very much! This helped me understand it easier than how my professor explained it.

Roxettna  Feb 19, 2019
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OK...got it now ????

CPhill  Feb 19, 2019
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Yes, I have. Thank you very much! ;)

Roxettna  Feb 19, 2019
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Let's look at (d) for another example

(f ° g ) (3)

We put 3 into g  and get g(3) = 2

Then...we put this result into f and get f(2)   = 16

Feb 19, 2019
#8
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Thanks Chris and Rom,

I always get confused by this notation too.

Feb 19, 2019
#9
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I always think that the "mathematicians" devise stuff like this to make math seem more "dark and deep" than it needs to be....also.....it sells a lot of math textbooks.....

CPhill  Feb 19, 2019
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Yea,

Set notation is the absolute worst.

Mathematicians like to have their own private  language.

I often think that 'mathematics' is my second language.  I am just not all that literate in it.   LOL.

Melody  Feb 19, 2019