Blog

Minimum value by Vivek Pandey

\color{blue}\bf{Problem}  Find the minimum value of \sin x+\cos x+\sin2x   over  \mathbb{R}

Solution: Let f(x)=sinx+cosx+sin2x , x\in\mathbb{R}.

Here the function {f} is continuous and differentiable on x\in\mathbb{R}.

Therefore f'(x)=\cos x−\sin x+2\cos2x=\cos x−\sin x+2(\cos2x−\sin2x)

  =(\cos x−\sin x)(1+2(\cos x+\sin x))

 

For critical points , f'(x)=0 which implies

\cos x=\sin x, \cos x+\sin x=−\frac12

 

The first equation gives x=\frac{\pi}{4},\frac{5\pi}{4}\in[0,2π]

The second equation gives \sin x_{0}+\cos x_{0}=−\frac12 or squaring we get

\sin2x_{0}=−\frac34, for some x_{0}.

Now we get the values of f at these critical points as

f(\frac{\pi}{4})=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}+\sin2\frac{\pi}{4}=1+\sqrt2.

f(\frac{5\pi}{4})=1−\sqrt2 and f(x_0)=−\frac54

Hence the least value of f is −\frac54

Leave a Reply

Your email address will not be published. Required fields are marked *