How would I sketch the following function.
$f(x)=\cos(x)+\sin(x)$ on $[0,2\pi]$
for my first derivative I got
$f'(x)=-\sin(x)+\cos(x)=0$
but How would I find the critical points I mean I know $\cos^{-1}=0$ can be 90 or 270 degrees but I am not sure if this is correct.
The second derivative is
$f''(x)=-\cos(x)-\sin(x)$
But how would I find the inflection point If I set the function zero.
Best Answer
For any $x\in [0,2\pi]$ you get $f'(x)=\cos (x) -\sin (x)$ and $f''(x)=-\sin (x)-\cos (x)$. Therefore $$\begin{align} f'(x)=0&\iff \cos (x) -\sin(x)=0\\ &\iff \cos (x) =\sin (x)\\ &\iff x\in \{\frac{\pi}{4}, \frac{5\pi}{4}\} \end{align}$$
Which gives you the critical points $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.
The last equivalence is easy to see geometrically if you look at the unit circle.
I don't understand why you're messing with $\cos ^{-1}$. Care to explain so we can help you?
Similarly, for $f''$ you'll get the zeros $\displaystyle \frac{3\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.