f(x) = sqrt (x-1)
g(x)= absolute (x+1)
pls find g(f(x)) and f(g(x))
thx
find g(f(x)) and f(g(x))?
2012-12-09 1:50 pm
回答 (5)
2012-12-09 2:02 pm
✔ 最佳答案
f(x) = √(x - 1)g(x) = |x + 1|
g(f(x)) = |√(x - 1) + 1|
= √(x - 1) + 1 answer//
f(g(x)) = √(|x + 1| - 1)
= √(x + 1 - 1)
= √(x) answer//
2016-08-03 10:35 pm
I would say go along with something's less expensive. If the SLI setup with 2 playing cards expenses the equal as a single card setup and each have precisely the identical performance then go for the single card. This at least offers you an option for going SLI with two high finish playing cards at some point. Additionally, you don't must buy a SLI motherboard for the single card, so if you haven't already bought one, this may retailer some huge cash.
2012-12-09 2:04 pm
Think back to early algebra and you did the following types of problems:
Say you have f(x) = 3x + 1
Then it was asked what f(2) was
What you did was ----- f(2) = 3*(2) + 1 = 7
And then they asked you what...say....f(a) was.
=> f(a) = 3*(a) + 1 = 3a + 1
So, here, you have f(x) and are asked for f(g(x)). It is no different here at all.
=> f(g(x)) = √( { |x + 1| } + 1 )....and this is about all you can do with this.
So.....hth.
Say you have f(x) = 3x + 1
Then it was asked what f(2) was
What you did was ----- f(2) = 3*(2) + 1 = 7
And then they asked you what...say....f(a) was.
=> f(a) = 3*(a) + 1 = 3a + 1
So, here, you have f(x) and are asked for f(g(x)). It is no different here at all.
=> f(g(x)) = √( { |x + 1| } + 1 )....and this is about all you can do with this.
So.....hth.
2012-12-09 1:56 pm
treat the inner function as the input of the outer function
g(f(x)) = | sqrt(x-1) +1 |
f(g(x) = sqrt( | x + 1| -1 )
g(f(x)) = | sqrt(x-1) +1 |
f(g(x) = sqrt( | x + 1| -1 )
2012-12-09 1:55 pm
f(x) = sqrt (x-1)
g(x)= absolute (x+1)
g(f(x)) = absolute( f(x) +1)
g(f(x) = | sqrt(x-1) + 1 |
f(g(x)) = sqrt( g(x) -1)
f(g(x) = sqrt( | x + 1| -1_
g(x)= absolute (x+1)
g(f(x)) = absolute( f(x) +1)
g(f(x) = | sqrt(x-1) + 1 |
f(g(x)) = sqrt( g(x) -1)
f(g(x) = sqrt( | x + 1| -1_
收錄日期: 2021-04-13 19:10:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121209055023AAHapu6