ktsconcepts ktsconcepts
  • 07-02-2020
  • Mathematics
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The range of F in F ( x ) = 2 sin x ; -pi/2< x< 2 is:

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kudzordzifrancis
kudzordzifrancis kudzordzifrancis
  • 14-02-2020

Answer:

[tex]-2\:<\:y\le2[/tex]

Step-by-step explanation:

The given function is [tex]f(x)=2sin(x)[/tex]

This function is has amplitude of 2.

This means that; [tex]-2\le f(x)\le 2[/tex]

Also the sine function will attain minimum at [tex]x=-\frac{\pi}{2}[/tex] and maximum is [tex]x=\frac{\pi}{2}[/tex].

Therefore the range is [tex]-2\:<\:y\le2[/tex]

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