Question

Prove that $f\left(x\right)=sinx+\sqrt{3}cosx$ has maximum value at $x=\frac{\pi }{6}.$

Answer 1

We have
$f\left(x\right)=sinx+\sqrt{3}cosx\therefore f^{\prime }\left(x\right)=cosx+\sqrt{3}(-sinx)for{f}^{\prime }\left(x\right)=0,cosx=\sqrt{3}sinx\Rightarrow tanx=\frac{1}{\sqrt{3}}=tan\frac{\pi }{6}\Rightarrow x=\frac{\pi }{6}$
Again, differentiating f'(x) we get
$f^{\prime \prime }\left(x\right)=-sinx-\sqrt{3}cosxAtx=\frac{\pi }{6},f^{\prime \prime }\left(x\right)=-sin\frac{\pi }{6}-\sqrt{3}cos\frac{\pi }{6}=-\frac{1}{2}-\sqrt{3}.\frac{\sqrt{3}}{2}=-\frac{1}{2}-\frac{3}{2}=-2<0$
Hence, at $x=\frac{\pi }{6}$, f(x) has maximum value. $\frac{\pi }{6}$ is the point of local maxima.