# Inc/dec/maxima/concavity

#### 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)

#### shenanigans87

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

#### TsAmE

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

#### shenanigans87

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.

#### skeeter

MHF Helper
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.

#### TsAmE

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)?