1. ## Inc/dec/maxima/concavity

f(x) = sinx + cosx, 0 <= x <= 2pi

a) Find the intervals on which f is increasing or decreasing.
b) Find the local maximum and minimum values of f
c) Find the intervals of concavity and the inflection points.

a)For f'(x) I got my critical points to be x = pi / 4 and x = 5pi / 4 but didnt know how to put this on a sign table in order to find when it is inc/dec

b) f(pi / 4) = (2)^1/2, but I didnt know how to work out
f(5pi / 4) = sin(5pi / 4) + cos(5pi / 4) without a calculator

c)For f''(x) my critical points were x = 3 pi / 4 and x = 3 pi / 2 and had the same problem as a)

2. Originally Posted by TsAmE
f(x) = sinx + cosx, 0 <= x <= 2pi

a) Find the intervals on which f is increasing or decreasing.
b) Find the local maximum and minimum values of f
c) Find the intervals of concavity and the inflection points.

a)For f'(x) I got my critical points to be x = pi / 4 and x = 5pi / 4 but didnt know how to put this on a sign table in order to find when it is inc/dec

b) f(pi / 4) = (2)^1/2, but I didnt know how to work out
f(5pi / 4) = sin(5pi / 4) + cos(5pi / 4) without a calculator

c)For f''(x) my critical points were x = 3 pi / 4 and x = 3 pi / 2 and had the same problem as a)
Make a graph of the function and you can easily see the answer to all these questions.

Here is a graph of the derivative.

If this were the function, you can clearly see that it is decreasing from 0->3pi/4, increasing from 3pi/4->7pi/4, decreasing from 7pi/4->2pi

local max is $\displaystyle (\frac{7\pi}{4},\sqrt{2})$
local min is $\displaystyle (\frac{3\pi}{4},-\sqrt{2})$

inflections at $\displaystyle (\frac{\pi}{4},0)$ and $\displaystyle (\frac{5\pi}{4},0)$

3. How would you do this without sketching the graph e.g using f'(x), f''(x) like I have done

4. Originally Posted by TsAmE
How would you do this without sketching the graph e.g using f'(x), f''(x) like I have done
Alright, consider this image:

What relation do you see between the critical points and the zeros of f'(x) and those of f(x)? This is your answer.

5. Originally Posted by TsAmE
How would you do this without sketching the graph e.g using f'(x), f''(x) like I have done
$\displaystyle f'(x) = \cos{x} - \sin{x}$

$\displaystyle f'(x) = 0$ at $\displaystyle x = \frac{\pi}{4}$ and $\displaystyle x = \frac{5\pi}{4}$

perform the first derivative test to locate extrema ...

in the interval $\displaystyle \left[0, \frac{\pi}{4}\right)$ , $\displaystyle f'(x) > 0$ ... $\displaystyle f(x)$ is increasing.

in the interval $\displaystyle \left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$ , $\displaystyle f'(x) < 0$ ... $\displaystyle f(x)$ is decreasing.

in the interval $\displaystyle \left(\frac{5\pi}{4}, 2\pi\right]$ , $\displaystyle f'(x) > 0$ ... $\displaystyle f(x)$ is increasing.

when f'(x) changes sign from positive to negative, f(x) has a relative maximum at that critical value.

when f'(x) changes sign from negative to positive, f(x) has a relative minimum at that critical value.

set f''(x) = 0 and perform the same analysis. inflection points on the graph of f(x) occur where f''(x) changes sign. f(x) will be concave up for f''(x) > 0 and concave down for f''(x) < 0.

all this basic info is in your text. look it over.

6. Oh I see makes a lot of sense, thanks for the explanation. Im just curious, you wouldnt be able to use a sign table if they didnt give you a closed interval right? e.g. for this question (without a closed interval) x < pi / 4 wouldnt be increasing (as it would soon decrease due to trig graphs oscillating infinitely)?