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Functions..... question below....
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\[g(x)=8x-x^2\] Find an expression, in terms of x, for \[g ^{-1}x\]
Write the function using y instead of f(x). Then switch x and y. Then solve for y. Then call y g^-1(x)
\[x=8y-y^2\] is that it please??
@mathstudent55
@amistre64
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that is correct enough yes; but in general, even powers have to have a modified inverse
\[g(x)=8x-x^2\] \[-g(x)=x^2-8x\] \[16-g(x)=x^2-8x+16\] \[16-g(x)=(x-4)^2\] \[\pm\sqrt{16-g(x)}=x-4\] \[4\pm\sqrt{16-g(x)}=x\] \[4\pm\sqrt{16-x}=g^{-1}(x)\]
notice how we would need to 2 functions, but of we restrict the domain to just real values 0 and above ... we get: 4+sqrt(16-x)
Thank you.....it took me so much time to realize what you have done....(:
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