Maximum/Minimum of $f(x)=\sin(2x)+5$ using second derivative test

Question

Find the maximum and minumum values of the function $f(x)=\sin(2x)+5$

It can easily be solved using the basic rules of inequalities. But, how do I solve it using first or second derivative test ?

My Attempt

$$ f'(x)=2\cos(2x)=0\implies2x=(2n+1)\frac{\pi}{2}\implies x=(2n+1)\frac{\pi}{4}\\\implies 2x=(4n+1)\frac{\pi}{2}\text{ or }2x=(4n+3)\frac{\pi}{2}\\ f''(x)=-4\sin(2x) $$

Is it the right way to approach the problem ?

Answer 1

$$ f(x)=\sin(2x)+5\\ f'(x)=2\cos(2x)\\ f''(x)=-4\sin(2x)\\ f'(x)=0\implies \cos(2x)=0\implies 2x=(2m+1)\frac{\pi}{2}\implies x=(2m+1)\frac{\pi}{4}\\ 2x=(2[2n]+1)\frac{\pi}{2}\text{ or }(2[2n+1]+1)\frac{\pi}{2}\\ \implies2x=(4n+1)\frac{\pi}{2} \text{ or }(4n+3)\frac{\pi}{2}\implies2x=2n\pi+\frac{\pi}{2}\text{ or }2n\pi+\frac{3\pi}{2}\\ \implies x=(4n+1)\frac{\pi}{4} \text{ or }(4n+3)\frac{\pi}{4}\implies2x=2n\pi+\frac{\pi}{4}\text{ or }2n\pi+\frac{3\pi}{4}\\ f''(2n\pi+\frac{\pi}{4})=-4\sin(2n\pi+\frac{\pi}{2})=-4\sin(\frac{\pi}{2})=-4<0\\ f''(2n\pi+\frac{3\pi}{4})=-4\sin(2n\pi+\frac{3\pi}{2})=-4\sin(\frac{3\pi}{2})=4>0\\ \implies x=2n\pi+\frac{\pi}{4} \text{is local maxima}\\ \implies x=2n\pi+\frac{3\pi}{4} \text{is local minima}\\ f_{max}=f(2n\pi+\frac{\pi}{4})=\sin(\pi/2)+5=1+5=6\\ f_{min}=f(2n\pi+\frac{3\pi}{4})=\sin(3\pi/2)+5=-1+5=4\\ $$

Answer 2

Yes, the derivative of f(x)= sin(2x)+ 5 is f'(x)= 2 cos(2x). That will be 0 when cos(2x)= 0. That happens, as you say, when 2x is an odd multiple of $\frac{\pi}{4}$- that is, $2x= (2k+ 1)\frac{\pi}{4}$ so $x= \frac{(2k+1)\pi}{8}$.

Yes, the second derivative is f''(x)= -4 sin(2x). When $x= \frac{(2k+1)\pi}{8}$ is $2x= \frac{(2k+1)\pi}{4}$. Those are odd multiples of $\frac{\pi}{4}$ sine of that goes "+, +, -, -".