a-7-1-3-9-1-3-b-4-Taqqoslang-

Question Number 129235 by enter last updated on 14/Jan/21

a = \sqrt[3]{7} + \sqrt[3]{9}

b = 4 T a q q o s l a n g

Commented by enter last updated on 14/Jan/21

c o m p a r e

Commented by enter last updated on 14/Jan/21

? ? ? ? ? ? ?

Answered by bemath last updated on 14/Jan/21

\sqrt[3]{7} + \sqrt[3]{9} = \frac{16}{\sqrt[3]{49} - \sqrt[3]{63} + \sqrt[3]{81}} < 4

Answered by mr W last updated on 13/Jun/21

f (x) = \sqrt[3]{x}

f^{'} (x) = \frac{x^{- \frac{2}{3}}}{3} > 0 f o r x > 0

f^{″} (x) = - \frac{2 x^{- \frac{5}{3}}}{9} < 0 f o r x > 0

t h a t m e a n s f o r x > 0, f (x) i s i n c r e a s i n g,

f^{'} (x) i s d e c r e a s i n g .

f (7) = \sqrt[3]{7} = \sqrt[3]{8 - 1} = \sqrt[3]{8} - \frac{f (8) - f (7)}{1}

f (9) = \sqrt[3]{9} = \sqrt[3]{8 + 1} = \sqrt[3]{8} + \frac{f (9) - f (8)}{1}

\frac{f (9) - f (8)}{1} < \frac{f (8) - f (7)}{1}

f (7) + f (9) = 2 \sqrt[3]{8} + [\frac{f (9) - f (8)}{1} - \frac{f (8) - f (7)}{1}]

< 2 \sqrt[3]{8}

\Rightarrow \sqrt[3]{7} + \sqrt[3]{9} < 2 \sqrt[3]{8} = 2 \times 2 = 4

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