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Minimum value

Problem. Find the minimum value of over Solution: Let Here the function is continuous and differentiable on Therefore For critical points , which implies , The first equation gives The second equation gives or squaring we get , for some Now we get the values of f at these critical points as and […]

Problem. Find the minimum value of sinx+cosx+sin2x over R
Solution: Let f(x)=sinx+cosx+sin2x ,x∈R
Here the function f is continuous and differentiable on R.
Therefore f′(x)=cosx−sinx+2cos2x=cosx−sinx+2(cos2x−sin2x) =(cosx−sinx)(1+2(cosx+sinx))
For critical points ,f′(x)=0 which implies cosx=sinx, cosx+sinx=−12
The first equation gives x=π4,5π4∈[0,2π]
The second equation gives sinx∘+cosx∘=−12 or squaring we get sin2x∘=−34, for some x∘
Now we get the values of f at these critical points as
f(π4)=sinπ4+cosπ4+sin2π4=1+2–√,
f(5π4)=1−2–√ and f(x∘)=−12−34=−54
Hence the least value of f is −54

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