1.Consider the function y = sin x on the interval [0,2pi]. (a) calculate the der

Question

1.Consider the function y = sin x on the interval [0,2pi].

(a) calculate the derivative, y'
(b) Identify the local maximum.
(c) Identify the local minimum
(d) Justify your responses for the local max and min by showing your work, or making a statement that summarizes your work.
(e) Identify interval(s) where the function is increasing
(f) identify intervals where the function is decreasing.
(g) Find the inflection point(s).
(h) Justify your reasoning for the above, either by showing your work, or making a statement that summarizes your work.
(i) Identify interval(s) where the function is concave up
(j) Identify interval(s) where the function is concave down.

2 .Find the inflection point(s) for the function y = csc x on [0,2pi] or argue, based on calculus, that the function does not have an inflection point.

Explanation / Answer

y = sin(x) over [0 , 2pi]

a) y' = d/dx(sinx) --> cosx ---> ANSWER

b) Local maximum :

First we find the critical points by doing y' = 0 and solving for x

cosx = 0

We must solve for x within [0 , 2pi]

cosx = 0

x = pi/2 , 3pi/2 ---> critical values

Now, y'' = -sin(x)

y'' at x = pi/2 is : -sin(pi/2) --> -1 ---> negative --> So, x = pi/2 is a MAXIMUM
y'' at x = 3pi/2 is : -sin(3pi/2) --> -(-1) --> 1 --> positive ---> So, x = 3pi/2 is a MINIMUM

At x = pi/2, y = sin(pi/2) --> y = 1
At x = 3pi/2, y = sin(3pi/2) --> y = -1

So, b) Local maximum = 1 when x = pi/2 ---> ANSWER

c) Local minimum = -1 when x = 3pi/2 ---> ANSWER

d) Since y'', the second derivative was negative at x = pi/2, x = pi/2 was a point for local maximum
And since y'' was positive for x = 3pi/2, x = 3pi/2 was a point of local minimum

e) Critical values were pi/2 , 3pi/2

And the interval was[0 , 2pi]

This splits the number line into [0 , pi/2) , (pi/2 , 3pi/2) and (3pi/2 , 2pi]

Region 1 : [0 , pi/2)
Testvalue = pi/4
y' = cos(x)
y' = cos(pi/4) --> sqrt(2)/2 ---> positive
So, INCREASING over this region

Region 2 : (pi/2 , 3pi/2)
Testvalue = pi
y' = cos(pi) = -1 --> negative
So, DECREASING over this region

Region 3 : (3pi/2 , 2pi]
Testvalue = 7pi/4
y' = cos(7pi/4) --> sqrt(2)/2 --> positive
So, INCREASING over this region

So, e) Increase : [0 , pi/2) U (3pi/2 , 2pi] ----> ANSWER

f) Decrease : (pi/2 , 3pi/2) ---> ANSWER

g) Inflection :

We found y'' = -sin(x) = 0

sinx = 0 over [0 , 2pi]

x = 0 , pi and 2pi are the inflection points

x = 0 and 2pi are also the endpoints of the given interval.
So, we do not count them when dealing with concavity

h) That split the number line as [0 , pi) and (pi , 2pi]

Region 1 : [0 , pi)
Testvalue = pi/2
y'' = -sin(x)
y'' = -sin(pi/2) = -1 ---> negative
So, CONCAVE DOWN over this region

Region 2 : (pi , 2pi]
Testvalue = 3pi/2
y'' = -sin(x)
y'' = -sin(3pi/2) = 1 --> positive
So, CONCAVE UP over this region

Notice that at x = pi, concave down becomes concave up
So, x = pi is an inflection point and by concavity, justification has been provided

i) Concave up : [0 , pi) ---> ANSWER

j) Concave down : (pi , 2pi] ---> ANSWER

-----------------------------------------------------------------------------------------------------------------------

2)

y = csc(x) over [0 , 2pi]

y' = first derivative = -csc(x)cot(x)

y'' = second derivative to be found using product rule

u = -csc(x) v = cot(x)
u' = csc(x)cot(x) , v' = -csc^2(x)

uv' + u'v

y'' = csc^3(x) + csc(x)cot^2(x)

And since we are finding inflection, we equate y'' = 0

csc^3(x) + csc(x)cot^2(x) = 0

csc(x) * (csc^2(x) + cot^2(x)) = 0

csc(x) = 0 and csc^2(x) + cot^2(x) = 0

csc(x) = 0 indicates that sin(x) = 1/0 --> sin(x) = undefined
But sinx is defined everywhere
So, cscx = 0---> no solution

csc^2x + cot^2x = 0
On the left, we have two perfect square terms ading to become equal to 0
Of the form a^2 + b^2 = 0, this never has any solution
So, csc^2x + cot^2x = 0 ---> no solution

So, there are no inflection points -----> ANSWER

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