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Maxima and minima

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Question

$The minimum value of 4 e^{2 x} + 9 e^{- 2 x} is$

Moderate

Solution

$\begin{array}{l} Let f (x) = 4 e^{2 x} + 9 e^{- 2 x} \\ ∴ f^{'} (x) = 8 e^{2 x} - 18 e^{- 2 x} \\ Put f^{'} (x) = 0 \Rightarrow 8 e^{2 x} - 18 e^{- 2 x} = 0 \\ e^{2 x} = 3 / 2 \Rightarrow x = \log (\sqrt{\frac{3}{2}}) \\ Again f^{''} (x) = 16 e^{2 x} + 36 e^{- 2 x} > 0 \\ Now f (\log {(3 / 2)}^{1 / 2}) = 4 e^{2 (\log {(3 / 2)}^{1 / 2})} + 9 e^{- 2 (\log {(3 / 2)}^{1 / 2}}) \\ = 4 \times \frac{3}{2} + 9 \times \frac{2}{3} = 6 + 6 = 12 \\ Hence, minimum value = 12. \end{array}$

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