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Mathematics 10 Online
OpenStudy (anonymous):

graph and find the inverse of f(x)=2x^2-4. Once you find the inverse, graph it too.

OpenStudy (zzr0ck3r):

This only has a restricted inverse since the function is not 1-1

OpenStudy (anonymous):

too bad it doesn't have an inverse ...

OpenStudy (zzr0ck3r):

and what I mean is that it does not have an inverse :)

OpenStudy (anonymous):

Hmm..so would I just graph f(x)=2x^2-4 and then say it doesnt have an inverse?

OpenStudy (zzr0ck3r):

You can say that by the definition of an inverse, it must be 1-1 and it is not.

OpenStudy (anonymous):

Thanks :)

OpenStudy (zzr0ck3r):

or I guess graph it and draw a horizontal line through any two points to show it is not 1-1

OpenStudy (anonymous):

What about y=-3x+6? I got -x+3/6 as the inverse

OpenStudy (zzr0ck3r):

hmm

OpenStudy (zzr0ck3r):

\(y=-3x+6\) Switch the \(x\) and the (y\) and solve for \(y\). \(x=-3y+6\\ x-6=-3y\\ \dfrac{x-6}{-3}=y\\ y=\dfrac{6-x}{3}\)

OpenStudy (anonymous):

not to butt in but you can still solve \[2y^2-4=x\]for \(y\) to find an inverse, it just won't be a function

OpenStudy (zzr0ck3r):

or restrict the domain on the first one

OpenStudy (anonymous):

add 4, divide by 2 and you get \[y^2=\frac{x+4}{2}\] but when you solve for \(y\) you get \[y=\pm\sqrt{\frac{x+4}{2}}\]

OpenStudy (anonymous):

the \(\pm\) make it not a function

OpenStudy (zzr0ck3r):

\(f:\mathbb{R}^+\rightarrow R, f(x) = 2x^2-4\) has as its inverse a proper function.

OpenStudy (zzr0ck3r):

I think they meant, use the graph to determine if it has an inverse.

OpenStudy (anonymous):

Thanks guys:D

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