Question

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

Answer 1

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.

Answer 2

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

Answer 3

Wouldn't the x value be 3?

Answer 4

nope, as the name says, \(f^{-1}(3)\) is the value of \(x\) that makes the function \(f(x)\) spit out \(3\)

Answer 5

\[f^{-1}(3) = a \implies f(a) = 3\]

so you need to solve \(f(a)=3\) for \(a\)

Answer 6

Ohhhh

Answer 7

\[3 = 3+x^2+\tan(\pi x/2)\]
\[0=x^2+\tan(\pi x/2)\]

you can eyeball the value of \(x\) that satisfies above equation

Answer 8

Yes!

Answer 9

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

Answer 10

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

Answer 11

np
what about part B

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Are you sure they cancel?

Answer 17

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

Answer 18

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 \(5\)