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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))

User Kingofmit
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1 Answer

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f(x) = 3x^(2)tan((\pi x)/(2))

f(x) = 3x^(2)[(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)[(1 - cos(\pi x))/(sin(\pi x))]

3 = 3y^(2)tan((\pi)/(2)y)

1 = y^(2)tan((\pi)/(2)y)

(1)/(tan((\pi)/(2)y)) = y^(2)

cot((\pi)/(2)y) = y^(2)

cot^(-1)[cot((\pi)/(2)y)] = cot^(-1)(y^(2))

(\pi)/(2)y = cot^(-1)(y^(2))

\pi y = 2cot^(-1)(y^(2))

y = (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)[(1 - cos(\pi x))/(sin(\pi x))]

5 = 3y^(2)tan((\pi)/(2)y)

1(2)/(3) = y^(2)tan((\pi)/(2)y)

(5)/(3tan((\pi)/(2)y)) = y^(2)

1(2)/(3)cot((\pi)/(2)y) = y^(2)

\sqrt{1(2)/(3)cot((\pi)/(2)y)} = y
User Nelson Auner
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