Find maximum and minimum value of sin2x+5 from second derivative test

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asked Nov 15, 2017 by Kumod kumar (55 points) 3

1 Answer

+2 votes
answered Nov 16, 2017 by Ankit Agarwal (28,847 points) 7 34 107

f(x)=sin 2x+5

We know that,

-1≤sinӨ≤1

-1≤sin2x≤1

Adding 5 on both sides,

 -1+5≤sin2x+5≤1+5

4≤sin2x+5≤6

Hence 

Max value of f(x)=sin2x+5 will be 6

Min value of f(x) =sin2x+5 will be 4

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