Can someone help😭 year 10 maths
F(x)=x^2-5/4
Find f^-1(19)
I got 9 but they are saying its 9 and -9 im so confused how do you get two answersss huhhh and please dont make it complicated
Find f^-1(19)
I got 9 but they are saying its 9 and -9 im so confused how do you get two answersss huhhh and please dont make it complicated
Reply 1
16
Original post
by Hrtz4rrF(x)=x^2-5/4
Find f^-1(19)
I got 9 but they are saying its 9 and -9 im so confused how do you get two answersss huhhh and please dont make it complicated
Find f^-1(19)
I got 9 but they are saying its 9 and -9 im so confused how do you get two answersss huhhh and please dont make it complicated
a function should only produce one output for a given input.
What exactly does the question say?
Reply 2
Original post
by Hrtz4rrF(x)=x^2-5/4
Find f^-1(19)
I got 9 but they are saying its 9 and -9 im so confused how do you get two answersss huhhh and please dont make it complicated
Find f^-1(19)
I got 9 but they are saying its 9 and -9 im so confused how do you get two answersss huhhh and please dont make it complicated
Because -9^2 and 9^2 both equal the same thing (81)
Reply 3
Hrtz4rr
OP
3
Original post
by isaac123444566because -9^2 and 9^2 both equal the same thing (81)
thank you so much
Reply 4
Original post
by Hrtz4rrthank you so much
nws!
Reply 5
square rooting can give both the positive and negative values. It's really annoying but you have to remember that
Reply 6
Original post
by maybeAverage_square rooting can give both the positive and negative values. It's really annoying but you have to remember that
you just need to know how negatives work when multiplied (i.e a negative * a negative = a positive)
Reply 7
21
Original post
by isaac123444566you just need to know how negatives work when multiplied (i.e a negative * a negative = a positive)
I guess the op should note davros's comment in #1. A (inverse) function can only return one value. So the two problems
x^2 = 81
x = sqrt(81)
where sqrt(...) is the usual inverse function, have different solutions. So the first is 9 and -9, whereas the second is only 9.
So really the wording of the question is important as well as who "they" are when talking about the answers being 9 and -9.
Reply 8
16
Original post
by isaac123444566Because -9^2 and 9^2 both equal the same thing (81)
Original post
by Hrtz4rrthank you so much
as mqb has expanded on , I was trying to get you to see that the question posted is meaningless - if f(x) sends 2 values of x to the same output, then f^-1(x) doesn't exist!
Either the domain of x needs to be restricted so that f^-1 only returns one value of x for the result of 19, or the question should have been written as "find the value(s) of x for which f(x) = 19" which does indeed have 2 solutions.
Did this question come from a book, or an exam, or a teacher?
Reply 9
To be even more pedantic, what does f^{-1} actually mean in your context? It's a very annoying notation that could mean two different things, but it's historic, so we don't care to fix it. I'll take that you mean "the inverse function of x"
As of this hour, if f(x)=x^2-5/4 (and supposedly defined over all real numbers), the inverse function f^{-1} doesn't even make sense, as davros stated. We need some domain on which f is defined, so that f is bijective. If you don't know what bijective means, pretend it's "strictly increasing/decreasing in the domain" for now. The key point is the inverse function, at the end of the day, is still a function - for every number from the domain, the function outputs one and only one number. If it outputs more than one number or no number, it's not a well-defined function to begin with.
As a sidenote, equally as annoying is "sin^{-1}(0)". This, we take sin^{-1} to be "arcsin", which actually outputs a specific range of values so that it makes sense (i.e. your calculator is correct). But as we know, sin(0)=sin(180)=sin(360)=sin(-180)=...=0.
-- downstairs is not even a thing in GCSE/AL, I think --
But if they mean f^{-1}({19}), now it becomes the pre-image of f of the singleton set {19}. This notation now makes sense, as it really means "what input does f outputs 19", and the answer is indeed the set {4.5, -4.5} (not "4.5 or -4.5", the answer should be a set).
As of this hour, if f(x)=x^2-5/4 (and supposedly defined over all real numbers), the inverse function f^{-1} doesn't even make sense, as davros stated. We need some domain on which f is defined, so that f is bijective. If you don't know what bijective means, pretend it's "strictly increasing/decreasing in the domain" for now. The key point is the inverse function, at the end of the day, is still a function - for every number from the domain, the function outputs one and only one number. If it outputs more than one number or no number, it's not a well-defined function to begin with.
As a sidenote, equally as annoying is "sin^{-1}(0)". This, we take sin^{-1} to be "arcsin", which actually outputs a specific range of values so that it makes sense (i.e. your calculator is correct). But as we know, sin(0)=sin(180)=sin(360)=sin(-180)=...=0.
-- downstairs is not even a thing in GCSE/AL, I think --
But if they mean f^{-1}({19}), now it becomes the pre-image of f of the singleton set {19}. This notation now makes sense, as it really means "what input does f outputs 19", and the answer is indeed the set {4.5, -4.5} (not "4.5 or -4.5", the answer should be a set).
(edited 11 months ago)
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