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find range: y= 1/2-sin3x

Note by Akash Sinha
3 years, 10 months ago

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$$y = \frac {1} {2 - sin 3x}$$ or $$y = \frac {1}{2} - sin 3x$$ ? · 3 years, 10 months ago

Anyway, for the first one, Range is $$(\frac {1}{3}, 1)$$

For second one, Range is $$(-\frac{1}{2}, \frac{3}{2})$$ · 3 years, 10 months ago

Reason: The min and max value of $$sin x$$ is $$-1$$ and $$+1$$ respectively. · 3 years, 10 months ago

How do we find min. and max. values ? · 3 years, 10 months ago

thanks · 3 years, 10 months ago