Given f(θ) = 2cos^2θ−4 sin θ on 0≤θ≤2π

Question

a) Find the intervals on which f is increasing or decreasing,

b) Find the local maximum and minimum values of f (use first derivative test)

c) Find the interval of concavity and inflection points

Answer 1

Start by finding the 1st derivative. If it is positive, then the value of the function is increasing. If negative, the function is DEcreasing.

When the 1st derivative goes to zero, it means the function is at a maximum or minimum.

"Find the interval of concavity and inflection points"---this is done using the second derivative:

https://www.dummies.com/education/math/calculus/how-to-locate-intervals-of-concavity-and-inflection-points/

Answer 2

a) Find the derivative and see where it is positive and where it is negative.

f(x) = 2cos^2(x) - 4sin(x)

So f'(x) = -4cos(x)sinx -4sin(x) = -4sin(x)(cosx +1)

On the interval 0< x < 2pi :

when is -4 >0?

when is -4<0?

when is sin(x)>0?

when is sin(x)<0

when is cos(x) + 1 >0

and when is cos(x) + 1<0

Now figure out when f'(x) > 0 (ie where f is increasing) and figure out where f'(x)<0 (ie where f is decreasing)

Answer 3

It's best to plot a few points graphically, to see where the function rises and falls. Then the answers will appear within the right general range needed to fit the graph.

Take the first derivative of f=2cos2x-4sinx

f'=2(-sin2x)(2)-4cosx=0 derivative of cos is -sin, derivative of sin is cos

-4sin2x=4cosx

-sin2x=cosx

sin2x=-cosx

2sinxcosx=-cosx sin2x formula sin2x=2sinxcosx

factor 2sinx=-1 and cosx=0 x=90 or 270, local minimums

sinx=-1/2

x=210 degrees or 330 degrees, 7pi/6 or 11pi/6 local maximums

These are 4 angles where the function has zero slope, where it changes from positive to negative at 210

and from negative to positive at 330. The zero slope means the function is at a local max or min

plug those angles back into the original function to get the values of the max & min

2cos2x-4sinx=2cos420-4sin210=2cos(14pi/6)-4sin(7pi/6)=2(1/2)-4(-1/2)=1+2=3 a local max

2cos660-4sin330=2(1/2)-4(-1/2)=1+2=3 a local max

local minimums are when x=90 or 270. substitute into original equation getting the values of the minimums: -6

find the 2nd derivative for concavity

also

plot a few points

At x=0, f=2 at x=360, f=2

x=180 f=2, x=90 f=-6

x=270 f=2

x=90 is a local minimum for f=-6 2cos180-4sin90=2(-1)-4(1=-6 f also =-6 at 270

f is decreasing from x=0 to x=90,

increasing from x=90 to x=180

decreasing x=180 to 270

increasing from 270 to 360

-4sin2x-4cosx is the first derivative

2nd derivative = -8cos2x+4sinx = 0

-2cos2x+sinx =0

-2(1-2sin²x)+sinx=0

-2+4sin²x+sinx=0

4sin²x+sinx-2=0 use quadratic formula

sinx=(-1+ or - sqr(1+32)/8

sinx=-1/8 + or - (33)^1/2/8 =-125 + or - .718 = -.843 or .583

x=arcsin(-.843) or arcsin(.583) = 35.66 and 144.34 degrees or -57.46 degrees=303.54 and 237.46

Those 4 angles are at the inflection points dividing concave from convex intervals of the function

The function is concave downward from 0 to 35.66

concave upward from 35.66 to 144.34

concave downward from 144.34 to 237.46

concave upward from 237.46 to 303.54

concave downward from 303.54 to 360