# Differential calculus: Extrema, INC/DEC, Concavity question?

Given the function y=(cosx)/(￼2+sinx), do the following:

i. Find the domain of f.

ii. Find the x-intercepts and y-intercept of the graph of f.

iii. Find where f is positive or negative.

iv. Find the intervals where f is increasing (↑), and the intervals where f is decreasing (↓).

v. Find all local maxima and all local minima of the function f. State how you know this, i.e. which test did you use?

vi. Find all intervals where the graph of f is concave up (∪) and all intervals where the graph of f is concave down (∩).

vii. Find all inflection points of the graph of f.

viii. Plot each “important point” accurately and then connect the dots by using the information that you

￼have gathered.

Thanks!

### 1 Answer

- Anonymous7 years agoFavorite Answer
This is a terribly long question to answer : I think you have to be more specific as to where you are stuck

i) the whole real line (the denominator is always >= 1 so there is no risk of dividing by zero)

ii) x intercepts : where cosx is zero, i.e. {π/2+nπ : n is an integer} y intercept : y(0)=1/2

Now, iii,iv and vi are all essentially the same question

f is positive exactly where f > 0

f is increasing exactly where f ' > 0

f is concave up exactly where f '' > 0

( and of course the opposite in each case when you replace > with < )

If you have answered iii, iv and and vi then you can immediately answer Vand vii

local maxima is wherever f ' changes from positive to negative as x increases (i.e. from left to right) In other words you have a local maxima wherever f changes from increasing to decreasing. Similarly, you have a local minima wherever f ' changes from negative to positive. Finding a local minima/maxima in this way is called using the First Derivative Test and it is a safe test in the sense that it always works. The Second Derivative Test doesn't always work.

you have an inflection point wherever f '' changes from negative to positive or from positive to negative. If f '' > 0 somewhere then it means f is concave up there, or loosely speaking, bending upwards. If f '' < 0 then f is concave down. So an inflection point is wherever the graph of f changes from concave up to concave down, and this can occur anywhere - sometimes on local minimas/maximas and sometimes not.

So : answer iii,iv and vi first from where you can then IMMEDIATELY read off the answer for v and vii.

I'll do one example : f > 0 where cos x > 0 since the denominator 2 + sinx >0 always

and the solution to cos x > 0 is {x : x ∈ (-π/2+t,π/2+t) where t=2nπ for some integer n}

In other words f > 0 on ...υ(-π/2-π,π/2-π)υ(-π/2,π/2)υ(-π/2+π,π/2+π)υ(-π/2+2π,π/2+2π)υ...