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for part A, you need to solve :
\[3 = 3+x^2+\tan(\pi x/2)\]

Wouldn't the x value be 3?

\[f^{-1}(3) = a \implies f(a) = 3\]
so you need to solve \(f(a)=3\) for \(a\)

Ohhhh

Yes!

so \(f(0) = 3 \implies f^{-1}(3) = 0\)

Wow so much simpler than I was doing it before! Thank you

np
what about part B

hold up,
we don't really need to find \(f^{-1}(5)\)

And then solve for the inverse and then place into the original

\(f\) eats \(f^{-1}\)

\[\require{cancel} \large { f(f^{-1}(5))\\~\\ \cancel{f}(\cancel{f^{-1}}(5))\\~\\5 }\]

Are you sure they cancel?

they dont cancel they generate new function
f(f^-1(x))=x

|dw:1439073410822:dw|

Ok

|dw:1439073520821:dw|

ofcourse assuming the inverse exists and one-to-one etc..

Ok that makes sense. Thank you for your help

np

plugin x = 0 and see that it evaluates to 0
i don't think there is a way to solve it analytically