SAT Math Part 8 - Identifying Undefined Function Values - By The Organic Chemistry Tutor
Transcript
00:00 | 29 for which value of X . Is the function | |
00:05 | shown below ? Undefined . Whenever you have a fraction | |
00:12 | or even a rational function , it's going to be | |
00:15 | undefined If the denominator is equal to zero . So | |
00:21 | therefore we can say that two x squared plus X | |
00:25 | -15 cannot equal zero . So let's factor this expression | |
00:33 | to factor it . We're going to multiply the leading | |
00:35 | coefficient by the constant term . So two times negative | |
00:40 | 15 is negative 30 . We need to numbers that | |
00:45 | multiply to negative 30 But add to the Middle Coefficient | |
00:48 | one . So two numbers that multiply to 30 would | |
00:53 | be six and 5 . Now we need to use | |
00:56 | negative five and positive six because these two numbers they | |
01:01 | multiply to negative 30 but add to positive one . | |
01:05 | So let's replace one . Acts with six X and | |
01:09 | negative five X . Yeah . Now let's factor by | |
01:15 | grouping in the first two terms , let's take out | |
01:18 | the G c f which is going to be two | |
01:19 | X two X squared divided by two X X . | |
01:24 | six x divided by two ax history . In the | |
01:28 | last two terms , let's take out the GCF . | |
01:31 | Which will be negative five negative five X divided by | |
01:34 | negative five X -15 divided by -5 . It's positive | |
01:40 | three . So next we need to factor out X | |
01:47 | plus story . And what goes in the second parentheses | |
01:51 | will be what we see here . two x -5 | |
01:55 | . So none of these factors must be equal to | |
01:57 | zero . So let's separate each factor . So for | |
02:10 | the first one , all we need to do is | |
02:11 | subtract both sides by three . So X can't be | |
02:14 | three . If it is the function will be undefined | |
02:18 | For the 2nd 1 , We need to add 5 | |
02:21 | to both sides and then we need to divide by | |
02:25 | two . So x cannot be 5/2 . So these | |
02:34 | are the values where the function is undefined . So | |
02:38 | be I mean that be But D . Is the | |
02:40 | answer Because five or 2 is listed -3 is not | |
02:44 | there number 30 . For which value of Why is | |
02:50 | the function shown below ? Undefined ? So the function | |
02:55 | will be undefined once again When the denominator is equal | |
02:59 | to zero . So why monastery squared plus five Y | |
03:06 | . Monastery minus 14 . We're going to set this | |
03:09 | to zero . So the function one exists or will | |
03:14 | be undefined when this portion is equal to zero . | |
03:18 | So we're gonna put this simple not equal to zero | |
03:21 | . Now , the way we're going to solve this | |
03:23 | is we're going to solve it by substitution . Let's | |
03:27 | substitute Why modest tree with a . So this becomes | |
03:31 | a squared plus five times a -14 . So now | |
03:38 | let's factory , We need to find two numbers that | |
03:42 | multiply to negative 14 but at the five . So | |
03:45 | this is going to be positive seven and negative two | |
03:50 | . So two factor it's going to be eight plus | |
03:51 | seven Times A -2 Sovereign for each factor . We're | |
03:59 | going to have that a can't be negative seven and | |
04:02 | A cannot be positive too . Mhm . So now | |
04:08 | at this point , what we need to do is | |
04:10 | replace A with y minus tree . So therefore Y | |
04:18 | -3 cannot equal -7 . and why -3 cannot equal | |
04:23 | to mm So let's add 3 to both sides . | |
04:32 | So we see that . Why can't be five ? | |
04:34 | Otherwise the function will be undefined . Now this answer | |
04:38 | is not listed For the next one . We need | |
04:40 | to add 3 to both sides . -3-plus 7 is | |
04:46 | negative four . So this is the one that we're | |
04:48 | looking for which is answer choice A . So negative | |
04:53 | four will cause the function to be undefined . So | |
04:59 | that's basically it for this problem . As you can | |
05:01 | see it's not that bad . But you need to | |
05:04 | know how to factor by substitution . |
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