Se-f-3-x-1-3-3-x-1-3-x-1-3-calcule-f-2-f-1-

Question Number 91681 by hmamarques1994@gmail.com last updated on 02/May/20

Se f (\sqrt{3^{\sqrt[3]{x}}} + 3^{\sqrt[3]{x}}) = \sqrt[3]{x}, calcule \frac{f (2)}{f (1)} \cdot

Commented by john santu last updated on 02/May/20

s e t \sqrt[3]{x} = t

f (3^{\frac{t}{2}} + 3^{t}) = t

(1) f (2); 2 = 3^{\frac{t}{2}} + 3^{t}, {(3^{\frac{t}{2}})}^{2} + 3^{\frac{t}{2}} - 2 = 0

3^{\frac{t}{2}} = 1 \Rightarrow t = 0, f (2) = 0

(2) f (1); {(3^{\frac{t}{2}})}^{2} + 3^{\frac{t}{2}} - 1 = 0

3^{\frac{t}{2}} = \frac{- 1 + \sqrt{5}}{2} \Rightarrow \frac{t}{2} = \log_{3} (\frac{\sqrt{5} - 1}{2})

t = \log_{3} (\frac{3 - \sqrt{5}}{2}) \Rightarrow f (1) = \log_{3} (\frac{3 - \sqrt{5}}{2})

(3) \frac{f (2)}{f (1)} = 0

Commented by hmamarques1994@gmail.com last updated on 02/May/20

S h o w, m a n!

Commented by john santu last updated on 02/May/20

m a n s h o w

Answered by mr W last updated on 02/May/20

s a y t = \sqrt{3^{\sqrt[3]{x}}} > 0

t^{2} = 3^{\sqrt[3]{x}}

f (t + t^{2}) = \log_{3} t^{2}

s a y u = t + t^{2}

t^{2} + t - u = 0

t = \frac{\sqrt{1 + 4 u} - 1}{2}

t^{2} = u - t = u - \frac{\sqrt{1 + 4 u} - 1}{2} = \frac{2 u + 1 - \sqrt{1 + 4 u}}{2}

f (u) = \log_{3} \frac{2 u + 1 - \sqrt{1 + 4 u}}{2}

\Rightarrow f (x) = \log_{3} \frac{2 x + 1 - \sqrt{1 + 4 x}}{2}

\Rightarrow \frac{f (2)}{f (1)} = \frac{\log_{3} \frac{5 - 3}{2}}{\log_{3} \frac{3 - \sqrt{5}}{2}} = \frac{0}{\log_{3} \frac{3 - \sqrt{5}}{2}} = 0

Commented by hmamarques1994@gmail.com last updated on 02/May/20

S h o w, m a n!