Let f(x)=3+×^2+tan((pi)(x)/2), where -1 <×<1 A) find f^-1 (3) B) find f (f^-1 (5))

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Let f(x)=3+×^2+tan((pi)(x)/2), where -1 <×<1 A) find f^-1 (3) B) find f (f^-1 (5))

Mathematics
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I know that when you inverse a function you switch the x and the y values. But I don't know what to do with the y value instead of x in tangent.
for part A, you need to solve : \[3 = 3+x^2+\tan(\pi x/2)\]
Wouldn't the x value be 3?

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nope, as the name says, \(f^{-1}(3)\) is the value of \(x\) that makes the function \(f(x)\) spit out \(3\)
\[f^{-1}(3) = a \implies f(a) = 3\] so you need to solve \(f(a)=3\) for \(a\)
Ohhhh
\[3 = 3+x^2+\tan(\pi x/2)\] \[0=x^2+\tan(\pi x/2)\] you can eyeball the value of \(x\) that satisfies above equation
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
Think of it like this : \(f^{-1}(5)\) is the value of \(x\) that makes \(f(x)\) spit out \(5\), so if you feed that value to \(f(x)\) again, it simply spits out \(5\)
|dw:1439073410822:dw|
Ok
|dw:1439073520821:dw|
ofcourse assuming the inverse exists and one-to-one etc..
Notice that in part B, you are inputting \(5\) to \(f^{-1}(x)\) (bottom box), it spits out \(x\), then you're inputting that value to \(f(x)\) again (top box) it spits out \(5\).
Ok that makes sense. Thank you for your help
np
I would like to know how to solve \(x^2 + tan (\pi x/2)=0\) . @rational please.
plugin x = 0 and see that it evaluates to 0 i don't think there is a way to solve it analytically
ok, I got you. Since It is trivial when we graph them out. I would like to know whether we can solve it algebraically or not. Thanks for clarifying.

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