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I am trying to find the intervals on which f is increasing or decreasing, local min and max, and concavity and inflextion points for $f(x)=\sin x+\cos x$ on the interval $[0,\pi]$.

I know at $\pi/4$ the derivative will equal zero. So that gives me my critical numbers, positive and negative $\pi/4$ so now I need to find the intervals which is not making any sense to me, I thought they could only change at critical numbers but $\pi$ and $2\pi$ are different values. I am getting a positive for $2\pi$ and a negtive for $\pi$. How can this happen if the only critical number is $\pi/4$?

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2 Answers 2

$f(x)=\sin(x)+\cos(x)$

$f'(x)=\cos(x)-\sin(x)$

critical points are when $f'(x)=0$:

i.e, at:

$\cos(x)=\sin(x)$ which can be satisfied by the values of x such as:

...,$-7{\pi}/4$ , $-3{\pi}/4$, ${\pi}/4$, $3{\pi}/4$,...

now, you need to examine the second derivative's sign at the above points:

$f''(x)=-\sin(x)-\cos(x)$

at $-7{\pi}/4$ , $f''(x)$ is (-) --> Local Max.

at $-3{\pi}/4$, $f''(x)$ is (+) --> Local Min.

at ${\pi}/4$ , $f''(x)$ is (-) --> Local Max.

at $3{\pi}/4$, $f''(x)$ is (+) --> Local Min.

The link (example) may help.

also, These plots may help: enter image description here

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It's a lovely graph. What did you do it in? –  mixedmath Oct 14 '11 at 1:42
    
Thanks for your comment. I used this tool: desmos.com/calculator –  Emmad Kareem Oct 14 '11 at 1:49
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In short, $\pi / 4$ is not the only critical value. In general, we know $\sin$ and $\cos$ are periodic, so we expect infinitely many critical values. We also know that if $\sin a = 0$, then $\sin (a + \pi) = 0$.

So when you take the derivative, you get $f'(x) = \cos x - \sin x$. You look for critical points - so you check when $\sin x = \cos x$, and you get when they're both positive (which you got), and when they're both negative (which you did not).

Note that $- \pi / 4$ is not a solution.

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Oh I missed a lot then. I can't think in terms of radians, degrees makes so much more sense. Anyways I think it is pi/4, 3pi/4, 5pi/4, 7pi/4. –  user138246 Oct 14 '11 at 1:12
    
@Jordan: For the purposes of calculus, you need to forget about degrees for the time being. They'll just inconvenience you... –  J. M. Oct 14 '11 at 1:14
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