Functions

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rachelmaddie

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12173D62-41B7-40CC-8D11-4A3C3846147B.png

Can someone please check my work?
Given functions
f(x) = 2x - 1
g(x) = x^2 - 2

(gof)(x) = g[f(x)] —>equation 1 (gof)(x) = g(2x - 1)
equation 2 = (2x - 1)^2 - 2
= 4x^2 + 1 - 4x - 2
= 4x^2 - 4x - 1

(gof)(x) = (2x - 1)^2 - 2 = 4x^2 - 4x + 1 - 2 = 4x^2
 
Where did the minus 4x go?

Why is 1 - 2 assumed equal to zero?

This not a perfect checking mechanism, but try putting in an example.

[MATH]f(3) = 2 * 3 - 1 = 5.\\ g(5) = 5^2 - 2 = 23 \text { but } 4 * 3^2 = 36[/MATH]The example does not work so the result is wrong.
 
Could the original question actually be asking for \(f(x) \cdot g(x) \)? I'm not sure what the "dot with a hole" means in the image!
 
I would personally assume that the image notation means g(x) * f(x) rather than g( f(x) ). But I could be mistaken. I would wait for verification from another helper unless you have a text book that clearly states what that notation in the image actually means. @JeffM do you know?
 
Cubist said:
Actually, this link https://mathmaine.com/2010/02/22/function-notation/ says that the open circle implies function of a function
Therefore g( f(x) ) is the correct interpretation. Please continue as of post#2, since JeffM is correct that there's a problem with your work.
Click to expand...
Yes. It is a notation that I do not like for students first studying functions because it does not build on the parentheses convention that the student already knows.
 
rachelmaddie said:
The result is 4x^2 - 4x - 1
Click to expand...
YES NOW CHECK IT. (It is best to check with two numbers other than 0 or 1, one positive and one negative, but you may not have time for that on a test.)

Let's pick three.

[MATH]f(3) = 2 * 3 - 1 = 6 - 1 = 5.[/MATH]
[MATH]g(5) = 5^2 - 2 = 25 - 2 = 23.[/MATH]
[MATH]4 * 3^2 - 4 * 3 - 1 = 4 * 9 - 12 - 1 = 36 - 13 = 23 \ \checkmark.[/MATH]
Try testing with minus 2.
 
The notation with the small circle was very popular in my students's time. But of course it can be very confusing for first timer. Even Spivak defines in his chapter of functions (Calculus book). 1594276527230.png
 
So, write it like this?
f(x) = 2x - 1
g(x) = x^2 - 2

(gof)(x) = g[f(x)] —>equation 1 (gof)(x) = g(2x - 1)
equation 2 = (2x - 1)^2 - 2
= 4x^2 + 1 - 4x - 2
= 4x^2 - 4x - 1

(gof)(x) = (2x - 1)^2 - 2 = 4x^2 - 4x + 1 - 2 = 4x^2 - 4x - 1
 
rachelmaddie said:
So, write it like this?
f(x) = 2x - 1
g(x) = x^2 - 2

(gof)(x) = g[f(x)] —>equation 1 (gof)(x) = g(2x - 1)
equation 2 = (2x - 1)^2 - 2
= 4x^2 + 1 - 4x - 2
= 4x^2 - 4x - 1

(gof)(x) = (2x - 1)^2 - 2 = 4x^2 - 4x + 1 - 2 = 4x^2 - 4x - 1
Click to expand...
It looks fine to me.
 
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