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

$f(x) = 3x^{2}tan(\frac{\pi x}{2})$
$f(x) = 3x^{2}[\frac{1 - cos(\pi x)}{sin(\pi x)}]$
$f(x) = 3x^{2}[csc(\pi x) - cot(\pi x)]$
$f(x) = 3x^{2}csc(\pi x) - 3x^{2}cot(\pi x)$
$y = 3x^{2}csc(\pi x) - 3x^{2}cot(\pi x)$
$x = 3y^{2}csc(\pi x) - 3y^{2}cot(\pi x)$

$x = 3y^{2}csc(\pi x) - 3y^{2}cot(\pi x)$
$3 = 3y^{2}csc(\pi x) - 3y^{2}cot(\pi x)$
$3 = 3y^{2}[csc(\pi x)] - 3y^{2}[cot(\pi x)]$
$3 = 3y^{2}[csc(\pi x) - cot(\pi x)]$
$3 = 3y^{2}[\frac{1 - cos(\pi x)}{sin(\pi x)}]$
$3 = 3y^{2}tan(\frac{\pi}{2}y)$
$1 = y^{2}tan(\frac{\pi}{2}y)$
$\frac{1}{tan(\frac{\pi}{2}y)} = y^{2}$
$cot(\frac{\pi}{2}y) = y^{2}$
$cot^{-1}[cot(\frac{\pi}{2}y)] = cot^{-1}(y^{2})$
$\frac{\pi}{2}y = cot^{-1}(y^{2})$
$\pi y = 2cot^{-1}(y^{2})$
$y = \frac{2cot^{-1}(y^{2})}{\pi}$

$x = 3y^{2}csc(\pi x) - 3y^{2}cot(\pi x)$
$5 = 3y^{2}csc(\pi x) - 3y^{2}cot(\pi x)$
$5 = 3y^{2}[csc(\pi x)] - 3y^{2}[cot(\pi x)]$
$5 = 3y^{2}[csc(\pi x) - cot(\pi x)]$
$5 = 3y^{2}[\frac{1 - cos(\pi x)}{sin(\pi x)}]$
$5 = 3y^{2}tan(\frac{\pi}{2}y)$
$1\frac{2}{3} = y^{2}tan(\frac{\pi}{2}y)$
$\frac{5}{3tan(\frac{\pi}{2}y)} = y^{2}$
$1\frac{2}{3}cot(\frac{\pi}{2}y) = y^{2}$
$\sqrt{1\frac{2}{3}cot(\frac{\pi}{2}y)} = y$