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- Thread starter Kwesi74
- Start date
Hello, Kwesi74!
Composite functions can be confusing at first . . .
I visualize it like this: It is the g-function "with the f-function inside".
. . . . . \(\displaystyle g(\underbrace{f(x)})\)
. . . . . . . . .\(\displaystyle \downarrow\) . . . . . . Replace \(\displaystyle f(x)\) with \(\displaystyle 2x - 1\)
. . . . .\(\displaystyle g(\overbrace{2x-1})\)
. . . . . . . . .\(\displaystyle \downarrow\) . . . . . . In the g-function, replace \(\displaystyle x\) with \(\displaystyle 2x - 1\)
. . . . . \(\displaystyle 4(2x-1) + 8\)
. . . . = .8x + 4
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Here's a baby-talk approach to composite functions.
We have a function, \(\displaystyle f(x) = 2x - 1\)
. . We can "process" numbers through this "machine".
. . . . . \(\displaystyle 3\;\rightarrow\;[\ f(x) = 2x - 1\ ]\;\rightarrow\;5\)
And we have another function, \(\displaystyle g(x) = 4x + 8\)
Suppose we put these two "machines" on an assembly line:
. . . . . \(\displaystyle ?\;\rightarrow\;[\ f(x)\ ]\;\rightarrow\;[\ g(x)\ ]\;\rightarrow\;?\)
If we insert a "3" in the first machine, we get:
. . . . . \(\displaystyle 3\;\rightarrow\;[\ f(x) = 2x-1\ ]\;\rightarrow\;5\;\rightarrow\;[\ g(x) = 4x + 8\ ]\;\rightarrow\;28\)
. . We put "3" into the first machine and get "5".
. . . . Then we put the "5" into the second machine and get "28".
The composite function, \(\displaystyle g(f(x))\), combines the two machines into one.
. . . . . . . . \(\displaystyle 3\;\rightarrow\;[\ g(f(x)) = 8x + 4\ ]\;\rightarrow\;28\)
Composite functions can be confusing at first . . .
Remember, this is <u>not</u> \(\displaystyle g(x)\cdot f(x),\) just multiplication. . This is: "g of the f-function".If \(\displaystyle f(x) = 2x-1\) and \(\displaystyle g(x) = 4x+8\), find \(\displaystyle g(f(x)).\)
I visualize it like this: It is the g-function "with the f-function inside".
. . . . . \(\displaystyle g(\underbrace{f(x)})\)
. . . . . . . . .\(\displaystyle \downarrow\) . . . . . . Replace \(\displaystyle f(x)\) with \(\displaystyle 2x - 1\)
. . . . .\(\displaystyle g(\overbrace{2x-1})\)
. . . . . . . . .\(\displaystyle \downarrow\) . . . . . . In the g-function, replace \(\displaystyle x\) with \(\displaystyle 2x - 1\)
. . . . . \(\displaystyle 4(2x-1) + 8\)
. . . . = .8x + 4
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Here's a baby-talk approach to composite functions.
We have a function, \(\displaystyle f(x) = 2x - 1\)
. . We can "process" numbers through this "machine".
. . . . . \(\displaystyle 3\;\rightarrow\;[\ f(x) = 2x - 1\ ]\;\rightarrow\;5\)
And we have another function, \(\displaystyle g(x) = 4x + 8\)
Suppose we put these two "machines" on an assembly line:
. . . . . \(\displaystyle ?\;\rightarrow\;[\ f(x)\ ]\;\rightarrow\;[\ g(x)\ ]\;\rightarrow\;?\)
If we insert a "3" in the first machine, we get:
. . . . . \(\displaystyle 3\;\rightarrow\;[\ f(x) = 2x-1\ ]\;\rightarrow\;5\;\rightarrow\;[\ g(x) = 4x + 8\ ]\;\rightarrow\;28\)
. . We put "3" into the first machine and get "5".
. . . . Then we put the "5" into the second machine and get "28".
The composite function, \(\displaystyle g(f(x))\), combines the two machines into one.
. . . . . . . . \(\displaystyle 3\;\rightarrow\;[\ g(f(x)) = 8x + 4\ ]\;\rightarrow\;28\)