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

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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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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