Harness.James

10-22-2008, 09:02 PM

yeah so this is a little confusing. What does f9g(x)) and g(f(x)) mean and how could I find it??

F(x) = -x^2 - 1 and G(x) = x + 5

Any Tips?

F(x) = -x^2 - 1 and G(x) = x + 5

Any Tips?

View Full Version : F(x) = -x^2 - 1, G(x) = x + 5; find f9g(x)) and g(f(x))

Harness.James

10-22-2008, 09:02 PM

yeah so this is a little confusing. What does f9g(x)) and g(f(x)) mean and how could I find it??

F(x) = -x^2 - 1 and G(x) = x + 5

Any Tips?

F(x) = -x^2 - 1 and G(x) = x + 5

Any Tips?

PAULK

10-22-2008, 11:15 PM

F(x) = -x^2 - 1 and G(x) = x + 5

Any Tips?

First tip: put some sandpaper on the shift key so you can find it quickly and it will come out as:

f(g(x)) and g(f(x))

which is what you meant, right?

Next tip: Don't be sloppy about capital letters and small letters. There will come a time when F(x) and f(x) refer to different things, and then you might find it confusing.

Now, then:

f(...) means to do the following:

1. Write the pattern for f(x) using empty parentheses instead of x:

f(x) = -x^2 - 1 becomes:

f( ) = - ( )^2 - 1

2. Take whatever you see inside the parentheses and copy and paste it:

f(g(x)) = - (g(x))^2 - 1

Then replace g(x) by its definition on the right side, work out the algebra, and you are done.

f(g(x)) = - (x + 5)^2 - 1

Any Tips?

First tip: put some sandpaper on the shift key so you can find it quickly and it will come out as:

f(g(x)) and g(f(x))

which is what you meant, right?

Next tip: Don't be sloppy about capital letters and small letters. There will come a time when F(x) and f(x) refer to different things, and then you might find it confusing.

Now, then:

f(...) means to do the following:

1. Write the pattern for f(x) using empty parentheses instead of x:

f(x) = -x^2 - 1 becomes:

f( ) = - ( )^2 - 1

2. Take whatever you see inside the parentheses and copy and paste it:

f(g(x)) = - (g(x))^2 - 1

Then replace g(x) by its definition on the right side, work out the algebra, and you are done.

f(g(x)) = - (x + 5)^2 - 1

Powered by vBulletin® Version 4.2.3 Copyright © 2017 vBulletin Solutions, Inc. All rights reserved.