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The minimum v...

Question

The minimum value of $\sqrt{e^{x^{2}} - 1}$ is

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Solution

$f (x) = \sqrt{(e^{x^{2}} - 1)} t h e s q u a r e r o o t t e r m s h o u l d a l w a y s b e p o s i t i v e o r z e r o . i t c a n n o t b e n e g a t i v e . s o e^{x^{2}} \geq 1 a n d w e w a n t t h e m i n i m u m v a l u e o f t h i s f (x) a n d t h a t c a n b e z e r o o n l y a t x = 0.$

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