1. ## Inflection And Concavity.

The directions say to find the points of inflection and discuss the concavity of the graph of the function.
The function is
$\displaystyle f(x)= sinx+cosx$ the intervals [0,2$\displaystyle pi$]
I know you supposed to take the derivative, then the 2nd then set the equation to zero
so i have up to
$\displaystyle 0= cosx-sinx$
Is this right?
if so what do i do next?

2. Originally Posted by Red350z
The directions say to find the points of inflection and discuss the concavity of the graph of the function.
The function is
$\displaystyle f(x)= sinx+cosx$ the intervals [0,2$\displaystyle pi$]
I know you supposed to take the derivative, then the 2nd then set the equation to zero
so i have up to
$\displaystyle 0= cosx-sinx$
Is this right?
if so what do i do next?
That's only the first derivative!!

The second derivative would be $\displaystyle f^{\prime\prime}\!\left(x\right)=-\sin x-\cos x$

To find inflection points, solve for x:

$\displaystyle -\sin x-\cos x=0\implies \sin x=-\cos x$

This is the case where $\displaystyle x=\frac{3\pi}{4},\,\frac{7\pi}{4}$.

Can you determine where the function is concave up or down now?

3. To get inflection points you have to solve f ''(x) = 0

You are solving f '(x) = 0 so you need to find f '' (x).
$\displaystyle -sinx - cos x = 0$

$\displaystyle -sinx = cosx$

This is only true when x = $\displaystyle \frac{3\pi}{4}$
x = $\displaystyle \frac{7\pi}{4}$

To the left of x = $\displaystyle \frac{3\pi}{4}$ f '' (x) < 0

To the right of x = $\displaystyle \frac{3\pi}{4}$ f '' (x) > 0

Therefore we have an ip( inflection point ) at x = $\displaystyle \frac{3\pi}{4}$

When $\displaystyle f''(x) <0$ Then f is concaving down
When $\displaystyle f''(x) >0$ then f is concaving up.

Do this for x = $\displaystyle \frac{7\pi}{4}$

4. Originally Posted by Chris L T521
That's only the first derivative!!

The second derivative would be $\displaystyle f^{\prime\prime}\!\left(x\right)=-\sin x-\cos x$

To find inflection points, solve for x:

$\displaystyle -\sin x-\cos x=0\implies \sin x=-\cos x$

This is the case where $\displaystyle x=\frac{3\pi}{4},\,\frac{7\pi}{4}$.

Can you determine where the function is concave up or down now?
I always get this wrong but hen f''(x) <0 Then f is concaving down, and when f''(x) >0 then f is concaving up.

5. Exactly.