Let f(x) = x2 - 16. Find f-1(x). (x2 means x to the 2 )
(f-1--> means f to the -1 )
\[f(x)=x^2-16\] and you want \[f^{-1}(x)\] right?
esactly
ok good tell the idiot who wrote this question that since \[y=x^2-16\] is a parabola, it is evidently not a one to one function so for example \[f(5)=f(-5)=9\] that there is no function \[f^{-1}(x)\]
on the other hand, we can still solve \[x=y^2-16\] for \(y\) we get \[x+16=y^2\\ \pm\sqrt{x+16}=y\]
Are you given a domain and range?
so i have no idea what you are supposed to put perhaps \[f^{-1}(x)=\pm\sqrt{x+16}\]
the \(\pm\) part tells you it is not a function
yea i think that's the answer
ask whoever wrote this question if they know what a function is, and if so, how you can have a function that is either plus or minus depending on your mood the answer is you cannot
well this question is from my exam of flvs
let me make a guess FLVS??
how did i know?
i keep asking for question id numbers from people who ask these questions' but no one ever wants to tell me
there questions are the most poorly written i have ever seen written by people who have no idea about math it is a sin you should insist on going to a real school, unless of course the schools in florida are actually even worse
*their
i left school because of bullying .....but hey can you help me with another one ?
sure
Let f(x) = -4x + 7 and g(x) = 10x - 6. Find f(g(x)). -40x+24 -40x + 31 -40x+64 -40x+70
ok we work from the inside out \[f(g(x))=f(10x-6)\] since \[\large f(\heartsuit)=-4\heartsuit+7\] you get \[f(10x-6)=-4(10x+6)-7\] now some algebra
ok i messed that up hold on
okay :)
\[f(10x-6)=-4(10x-6)+7\]thats better
now algebra \[-40x+24+7\\ -40x+31\]
go with B
got more or is that it?
if you wanna keep helping me i have 2 more
sure no problem
Let f(x) = 12 over the quantity of 4 x + 2. Find f(-1).|dw:1417017843604:dw|
\[y=\frac{12}{4x+2}\]
switch x and y \[x=\frac{12}{4y+2}\] and solve for \(y\)
\[x=\frac{12}{4y+2}\\ x(4y+2)=12\] \[4xy+2x=12\\ 4xy=12-2x\\ y=\frac{12-2x}{4x}\]
other in answers include \[f^{-1}(x)=\frac{6-x}{2x}\]
this is a little bit complicated
did you see an answer that looked like \[f^{-1}(x)=\frac{6-x}{2x}\]
yes
-6 -2 2 6 I'm guessing is 6
oh wait, all you need is \(f(-1)\)?
\[f(-1)=\frac{12}{4\times (-1)+2}=\frac{12}{-2}=-6\]
go with \(-6\)
phh okay thank you
next?
Let f(x) = 2x + 2. Solve f-1(x) when x = 4. 1 3 4 10
this one is quick solve \[2x+2=4\] for \(x\) in two steps
\[2x+2=4\\ 2x=2\\ x=1\]
i got that one right , i did it on my own but i wanted to see if i was right thank you
yw
that's all thank you for your help
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