What is the $x$-coordinate for the point of inflection where the graph of $f(x)$ goes from concave down to concave up? $$f(x)=\sin(x)+\cos(x)\; \text{when}\;0\le x\le2$$ What is the $x$-coordinate for the point of inflection where the graph of $f(x)$ goes from concave down to concave up?
Please explain as much as possible. Also include the answer if you want to be chosen as best answer. I haven't been able to solve this for a week. Thank you.
 A: As has been pointed out there is no solution if you impose the requirement $0<x<2$. However, if you do not have this restriction, read ahead.
You want the second derivative to be zero and the next higher nonzero derivative should be an odd derivative.
$$f'' = -\sin(x)-\cos(x) = 0$$
Gives
$$ \tan x= -1 \Rightarrow x= 3 \pi/4$$
It is clear that $f''' \neq 0$
BTW, we are not here to have our answers chosen. We are here to help. So that statement in your question is really an insult. Frankly, all of us do this because we want to help.
A: The change of concavity corresponds to a point where the second derivative of the function vanishes. In your case,   
f(x) = sin(x) + cos(x)
f'(x) = cos(x) - sin(x)
f''(x)=-sin(x) - cos(x)  
The second derivative cancels where sin(x) = - cos(x), that is to say tan(x) = -1; for x > 0, this corresponds to x = 3 Pi/4 which is larher than 2. So, as a plot of your function would easily show, for 0 < x < 2, there is no inflection point.   
If you compute the value of the second derivative at any point in the range, say x = 0, you notice that the second derivative is negative. So your curve is concave down.  
For sure, as user44197 points it out, you must also be sure that the third derivative is not zero at that point.
A: $f(x)=\sin x +\cos x = \sqrt{2}\cos ( x-\pi/4)$, $f'(x)=-\sqrt{2}\sin(x-\pi/4)$
$f''(x)=-\sqrt{2}\cos (x-\pi/4)$ 
$f''(x)=0$ 
$\cos (x-\pi/4)=0$ there is no solution in $[0,2]$ there is no nflection point.
