f n ( x ) = f 1 ( x ) f^{n}(x) = f^{-1}(x)

Calculus Level pending

Let n n be a positive integer and b > n 1 b > n - 1 , f ( x ) = a x b f(x) = ax^{b} and f n ( x ) = f 1 ( x ) f^{n}(x) = f^{-1}(x) , where f n ( x ) f^{n}(x) is the n n th derivative of f ( x ) f(x) .

(1) Find the area of the region bounded by f ( x ) f(x) and f 1 ( x ) f^{-1}(x) as a function of n n .

(2) Using n = 3 : n = 3: If the area A A of the region R R above bounded by f ( x ) f(x) and f 1 ( x ) = f 3 ( x ) f^{-1}(x) = f^{3}(x) can be expressed as A = ( a b c ) ( d a + d b a ) a d ( b + c b + d ) A = (\dfrac{a}{\sqrt{b} - c}) * (\dfrac{d^a + d\sqrt{b}}{a})^{\dfrac{a}{d}} * ( \dfrac{\sqrt{b} + c}{\sqrt{b} + d}) \:\ , where a , b , c a,b,c and d d are coprime positive integers, find a + b + c + d a + b + c + d .


The answer is 19.

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1 solution

Rocco Dalto
Dec 28, 2019

f ( x ) = a x b f n ( x ) = a m x b n f(x) = ax^{b} \implies f^{n}(x) = am x^{b - n} where m = b ( b 1 ) ( b 2 ) ( b n + 1 ) m = b(b - 1)(b - 2) *** (b - n + 1)

and f 1 ( x ) = 1 a 1 b x 1 b f^{-1}(x) = \frac{1}{a^{\frac{1}{b}}} x^{\frac{1}{b}}

For x b n = x 1 b b n = 1 b b = n + n 2 + 4 2 x^{b - n} = x^{\frac{1}{b}} \implies b - n = \dfrac{1}{b} \implies b = \dfrac{n + \sqrt{n^2 + 4}}{2} for b > n 1 b > n - 1 .

For a m = 1 a 1 b = 1 a b n am = \dfrac{1}{a^{\dfrac{1}{b}}} = \dfrac{1}{a^{b - n}} \implies a b n + 1 m = 1 a = ( 1 m ) 1 b n + 1 a^{b - n + 1} m = 1 \implies a = (\dfrac{1}{m})^{\dfrac{1}{b - n + 1}}

f ( x ) = ( 1 m ) 1 b n + 1 x b \implies f(x) = (\dfrac{1}{m})^{\dfrac{1}{b - n + 1}} * x^{b}

and

f n ( x ) = m b n b n + 1 x b n = f 1 ( x ) f^{n}(x) = m^{\frac{b - n}{b - n + 1}} x^{b - n} = f^{-1}(x)

f ( x ) = f 1 ( x ) x b n ( m b n b n + 1 ( 1 m ) 1 b n + 1 x n ) f(x) = f^{-1}(x) \implies x^{b - n}(m^{\frac{b - n}{b - n + 1}} - (\dfrac{1}{m})^{\frac{1}{b - n + 1}} x^{n})

x = 0 x = 0 and for x 0 x = m 1 n x \geq 0 \implies x = m^{\frac{1}{n}} for case n n even.

\implies The area A = 0 m 1 n f 1 ( x ) f ( x ) d x = A = \displaystyle\int_{0}^{m^{\frac{1}{n}}} f^{-1}(x) - f(x) \:\ dx = ( 1 b n + 1 ) m 2 n ( 1 b ( b n + 1 ) ) m 2 n (\dfrac{1}{b - n + 1})m^{\frac{2}{n}} - (\dfrac{1}{b(b - n + 1)})m^{\frac{2}{n}}

= ( 1 b n + 1 ) ( m 2 n ) ( b 1 b ) = = (\dfrac{1}{b - n + 1})(m^{\frac{2}{n}})(\dfrac{b - 1}{b}) =

( 1 b n + 1 ) ( b ( b 1 ) ( b 2 ) ( b n + 1 ) ) 2 n ( b 1 b ) \boxed{(\dfrac{1}{b - n + 1})(b(b - 1)(b - 2) * * * (b - n + 1))^{\dfrac{2}{n}}(\dfrac{b - 1}{b})}

Note: You could have used the Gamma function if so desired.

For n = 3 n = 3 we have:

b = 3 + 13 2 b = \dfrac{3 + \sqrt{13}}{2} \implies The area A = A = ( 2 13 1 ) ( 9 + 3 13 2 ) 2 13 ( 13 + 1 13 + 3 ) = (\dfrac{2}{\sqrt{13} - 1}) (\dfrac{9 + 3\sqrt{13}}{2})^{\dfrac{2}{13}} (\dfrac{\sqrt{13} + 1}{\sqrt{13} + 3}) =

( 2 13 1 ) ( 3 2 + 3 13 2 ) 2 3 ( 13 + 1 13 + 3 ) = (\dfrac{2}{\sqrt{13} - 1}) * (\dfrac{3^{2} + 3\sqrt{13}}{2})^{\dfrac{2}{3}} * (\dfrac{\sqrt{13} + 1}{\sqrt{13} + 3}) =

( a b c ) ( d a + d b a ) a d ( b + c b + d ) a + b + c + d = 19 (\dfrac{a}{\sqrt{b} - c}) * (\dfrac{d^a + d\sqrt{b}}{a})^{\dfrac{a}{d}} * ( \dfrac{\sqrt{b} + c}{\sqrt{b} + d}) \implies a + b + c + d = \boxed{19} .

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