## anonymous one year ago 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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1. anonymous

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.

2. rational

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

3. anonymous

Wouldn't the x value be 3?

4. rational

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

5. rational

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

6. anonymous

Ohhhh

7. rational

$3 = 3+x^2+\tan(\pi x/2)$ $0=x^2+\tan(\pi x/2)$ you can eyeball the value of $$x$$ that satisfies above equation

8. rational

Yes!

9. rational

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

10. anonymous

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

11. rational

12. rational

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

13. anonymous

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

14. rational

$$f$$ eats $$f^{-1}$$

15. rational

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

16. anonymous

Are you sure they cancel?

17. ikram002p

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

18. rational

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 $$5$$

19. rational

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20. anonymous

Ok

21. rational

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22. rational

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

23. rational

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 $$5$$.

24. anonymous

Ok that makes sense. Thank you for your help

25. rational

np

26. Loser66

I would like to know how to solve $$x^2 + tan (\pi x/2)=0$$ . @rational please.

27. rational

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

28. Loser66

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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