# Analysis of a function's concavity and increase/decrease without a graphing calc

The function is: $$f(x)= \sin(x)-\cos(x)$$ over the interval $$[-\pi, \pi]$$.

To evaluate the increasing / decreasing portion of the function I look at where the first derivative is $$>0$$. $$f'(x)=\cos(x) + \sin(x)$$. I want to know where this is $$>0$$ without a graphing calculator.

I'm stuck here. I know that the derivative would be greater than zero in Quadrant 1. I know that the derivative would be $$<0$$ in Q3. I can intuitively think of where in Q2 and Q3 that the derivative would be $$>0$$, but I don't know how to show this mathematically. I also don't think I totally understand the significance of the interval, since the interval is essentially over the entire $$0-2\pi$$ range but stated in this unusual manner.

• Have you tried $f^\prime(x)=0$? Because $f^\prime$ is continuous, it must have a 0 between any positive and negative values. That means that you can the interval between zeros by picking any point in them and finding if it's positive or negative. – memerson Oct 23 '18 at 20:16
• I think part of your second sentence is missing. Can you complete. – user163862 Oct 23 '18 at 20:25
• Let's take an example. Say $g(x)$ is continuous, $g(x)=0$ at $x=1,5$ and the sign of $g$ on the interval $[0,10]$. Let's look at $[0,1)$. Let $x,y\in[0,1)$ with $x<y$. Suppose that one of them is positive and the other is negative. By the intermediate value theorem, there is some $z\in [x,y]$ where $g(z)=0$. But that would give us another $0$, so $x$ and $y$ must have the same sign. As such, for every $x\in[0,1)$, $g(x)$ has the same sign, so if we know that $g(.5)$ is postive, so is the rest of $[0,1)$. The same applies to $(1,5)$ and $(5,10]. Does that make sense? – memerson Oct 23 '18 at 20:33 ## 1 Answer $$\cos x+\sin x=\sqrt{2}\sin(x+\pi/4)$$ See the trigonometric identity here: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations • This is extremely useful. However, when I set this identity$>0$I end up with – user163862 Oct 23 '18 at 20:43 • When I solve for the inverse$sin$of$0$, if I use their boundaries I miss the answer of$-\pi/4$. If I used the typical boundaries of$0$to$2 \pi\$ I get the book answers. Thank you! – user163862 Oct 23 '18 at 20:53
• I see this now. – user163862 Oct 23 '18 at 21:10