Maxima and minima

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

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

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Maxima and minima

Remember concepts with our Masterclasses.

The minimum value of 4e2x+9e-2x is

Let f(x)=4e2x+9e−2x∴f'(x)=8e2x−18e−2x Put f'(x)=0⇒8e2x−18e−2x=0e2x=3/2⇒x =log32 Again f''(x)=16e2x+36e−2x>0 Now flog(3/2)1/2=4e2log(3/2)1/2+9e−2log(3/2)1/2=4×32+9×23=6+6=12 Hence, minimum value =12.

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