# Thread: Help with local max/min, concavity, increasing decreasing?

1. ## Help with local max/min, concavity, increasing decreasing?

13) $f(x)=sin x + cos x 0\leq x \leq 2\pi$
$f'(x)=cosx-sinx = 0$
$x = \frac{\pi}{4} and \frac{5\pi}{4}$
how do I find where the function increase and decrease?
If I put these into a sign chart, I end up with all 0's
also how do I find the concavity?
I get that $f''(x) = -sin x - cos x$
so $x = \frac{7\pi}{4} and \frac{3\pi}{4}$
and for this also, if I plug it into f''(x) for a sign chart, the results will still be 0
is there another method for trig to do these problems?

2. Originally Posted by dorkymichelle
13) $f(x)=sin x + cos x 0\leq x \leq 2\pi$
$f'(x)=cosx-sinx = 0$
$x = \frac{\pi}{4} and \frac{5\pi}{4}$
how do I find where the function increase and decrease?
$x = \frac{\pi}{4} and \frac{5\pi}{4}$ are split points that split up the domain into three intervals

$0\leq x\leq \frac{\pi}{4}$

$\frac{\pi}{4}\leq x\leq \frac{5\pi}{4}$ and

$\frac{5\pi}{4}\leq x\leq 2\pi$

Originally Posted by dorkymichelle
If I put these into a sign chart, I end up with all 0's
To determine the sign of $f'(x)$ in each of these intervals use a value of x in the interior of the interval

To determine the sign of $f'(x)$ in the interval, $0\leq x\leq \frac{\pi}{4}$, for example, you could use $x=\frac{\pi}{6}$

$f'(\frac{\pi}{6})=\cos{\frac{\pi}{6}}-\sin{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}-\frac{1}{2}>0$

If $f'(x)>0$, then f(x) is increasing in the interval

If $f'(x)<0$, then f(x) is decreasing in the interval

Determine the sign of $f'(x)$ in the other two intervals

Originally Posted by dorkymichelle
also how do I find the concavity?
I get that $f''(x) = -sin x - cos x$
so $x = \frac{7\pi}{4} and \frac{3\pi}{4}$
$x = \frac{7\pi}{4} and \frac{3\pi}{4}$ are split points for the second derivative

They split up the domain into the intervals

$0\leq x\leq \frac{3\pi}{4}$

$\frac{3\pi}{4}\leq x\leq \frac{7\pi}{4}$ and

$\frac{7\pi}{4}\leq x\leq 2\pi$

Determine the sign of $f"(x)$ in each interval

If $f"(x)>0$, then f(x) is concave up

If $f"(x)<0$, then f(x) is concave down

3. Oh I see, I guess I just had a "trig scares me" moment and forgot to find a value between those intervals and used the crit. points instead.