How do I setup this function?

typingken

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Feb 3, 2006
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How do I setup this function?
I am not even sure what the circle stands for?
Given that f(x)=6 x + 2 and g(x)=7 x - 7, calculate
I am not even sure how to plug these in? Please help me out here!

(a) f\circle g(x)=
(b) g\circle f(x)=
(c) f\circle f(x)=
(d) g\circle g(x)=
 
typingken said:
I am not even sure what the circle stands for?
If you haven't covered function composition, it's going to be difficult to explain it here. (I'm afraid we're just not set up to teach courses in this environment. Sorry.) In brief, (f o g)(x) = f(g(x)), so you plug the expression for g(x) in for all the x's in f(x). Foir instance:

. . .f(x) = 3x + 2, g(x) = x<sup>2</sup> - 1

. . .(f o g)(x) = f(g(x)) = f(x<sup>2</sup> - 1)
. . . . . . . . . .= 3(x<sup>2</sup> - 1) + 2
. . . . . . . . . .= 3x<sup>2</sup> - 3 + 5
. . . . . . . . . .= 3x<sup>2</sup> - 1

. . .(g o f)(x) = g(f(x)) = g(3x + 2)
. . . . . . . . . .= (3x + 2)<sup>2</sup> - 1
. . . . . . . . . .= 9x<sup>2</sup> + 12x + 4 - 1
. . . . . . . . . .= 9x<sup>2</sup> + 12x + 3

As you can see, the order in a composition can be very important. :wink:

Eliz.
 
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