Composite Functions: f(g(x)) and g(f(x)) | MathHelp.com - Free Educational videos for Students in K-12

Composite Functions: f(g(x)) and g(f(x)) | MathHelp.com - By MathHelp.com

Transcript


00:0-1	in this example were given the functions F of X
00:03	equals three x minus two , and G of X
00:07	equals root X . And we're asked to find the
00:10	composite functions F of G of nine and G of
00:15	F of nine . To find F of G of
00:18	nine . We first find G of nine since G
00:23	of X equals root X . We can find G
00:26	of nine by substituting a nine in for the X
00:30	. In the function to get G of nine equals
00:34	route nine and the square root of nine is 3
00:38	, So G of nine equals 3 . Now ,
00:43	since G of nine equals 3 , F of G
00:46	of nine is the same thing as F of three
00:50	. So our next step is to find f of
00:53	three and remember that F of X equals three X
00:57	minus two . So to find F of three ,
01:01	we substitute a three in for the X in the
01:04	function And we have f of three equals 3 times
01:09	3 -2 . Finally three times 3 -2 simplifies to
01:16	9 -2 or seven . So f of three equals
01:20	7 . Therefore f of G of nine equals 7
01:26	. Next to find G of F of nine .
01:29	We first find f of nine since F of X
01:34	equals three x minus two . We find F of
01:37	nine by substituting a nine in for the X .
01:41	In the function . to get f of nine equals
01:45	3 times 9 -2 , Which simplifies to 27 -2
01:52	or 25 . So f of nine equals 25 .
01:57	Now , since F of nine equals 25 G of
02:01	F of nine is the same thing as G of
02:04	25 . So our next step is to find g
02:08	of 25 and remember that G of X equals rude
02:12	X . So to find G of 25 we substitute
02:16	a 25 in for the X . In the function
02:20	To get G of 25 equals route 25 . Finally
02:26	, the square root of 25 is five , So
02:29	GF 25 equals five . Therefore , Gff of nine
02:34	equals 5 . It's important to recognize that the value
02:39	of gff of nine , which is five , is
02:42	different from the value of F of G of nine
02:46	, which is seven .

Summarizer

DESCRIPTION:

In this problem, weâre asked to add the given polynomials, then weâre asked to subtract the second polynomial from the first. In part a, to add the given polynomials, we simply add parentheses t^2 + 6t â 9 + parentheses t^2 + 7t - 3. Notice that I used parentheses around the polynomials. This is a good habit to get into, even though the parentheses will not affect the addition. Next, we simply add the like terms, t^2 + t^2 is 2t^2, 6t + 7t is 13t, and -9 - 3 is -12. So we have 2t^2 + 13t â 12. In part b, weâre asked to subtract the second polynomial from the first, so we have parentheses t^2 + 6t â 9 minus parentheses t^2 + 7t - 3. Notice that the second polynomial is subtracted from the first. And again, notice that we use parentheses around each polynomial. Now, itâs important to understand that the minus sign outside the second set of parentheses can be thought of as a negative 1, so we need to distribute the -1 through each of the terms in the second set of parentheses. So, after rewriting our first polynomial, t^2 + 6t â 9, we have -1 times t^2, or ât^2, -1 times positive 7t, which is -7t, and -1 times -3, which is positive 3. Now, we combine like terms. t^2 â t^2 cancels out, positive 6t minus 7t is -1t, or ât, and -9 + 3 is -6. So we have ât â 6. Makes sure to distribute the negative 1 through the parentheses when subtracting the second polynomial from the first.

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