Calculus homework help: Find the intervals on which y=sin(x) is increasing, decreasing, concave up/down?

Read your textbook.

Learn the First Derivative Test -- This determines when your function is increasing (first derivative is positive) and when your function is decreasing (first derivative is negative).

Learn the Second Derivative Test -- This determines when your function is concave up and when it is concave down.

Learn inflection points are where concavity changes.

Extreme values happern where what changes? (In part e, you want to include the x and the f(x) values when you hit maximima and minima. In part f, you want to include x and f(x) values.)

For this question, it will be best if you draw out y=sin(x) from 0 to 2pi and observe it.

a)

You will notice that y increases from 0 to pi/2 and from 3pi/2 to 2pi.

However, the domain of y =sin(x) is from negative infinity to infinity, since the question did not state what is the interval y = sin(x) is in (or you did not type it out).

Hence, the correct answer will be:

2*n*pi < x < pi/2 + 2*n*pi, where n is any integer

2*n*pi + 3pi/2 < x < 2pi + 2*n*pi, where n is any integer

b)

Look at your graph from 0 to 2pi again.

The graph is decreasing from pi/2 to 3pi/2

Thus, the standard answer will be:

pi/2 + 2*n*pi < x < 3pi/2 + 2*n*pi, where n is any integer

c)

For sine graphs, it is easy to notice where the graph concaves upwards and where it concaves downwards. The part where it forms a U-shape, otherwise known as the happy face, is where it concaves upwards. The part where it forms a n-shape, otherwise known as the sad face, is where it concaves downwards.

Looking at your graph, you will notice that a happy face is formed when pi < x < 2pi

Thus, this is where y = sin(x) concaves upwards.

Hence, the general solution will be:

pi + 2*n*pi < x < 2pi + 2*n*pi, where n is any integer

d)

The unhappy face, on the otherhand, is formed when 0 < x < pi

Thus, this is where y = sin(x) concaves downwards.

Hence, the general solution will be:

2*n*pi < x < pi + 2*n*pi, where n is any integer

e)

Yes, 1 and -1 are both the local and absolute extreme values.

However, if the question is asking you where they occur at, then it is another thing altogether.

Looking at the graph you drawn again, you can see that f(x) = 1 occurs when x = pi/2 and f(x) = -1 occurs when x = 3pi/2

Thus, the local extreme values occur at:

pi/2 + 2*n*pi, where n is any integer

3pi/2 + 2*n*pi, where n is any integer,

which we can simplify to:

pi/2 + n*pi, where n is any integer

f)

The inflection points are where the graph changes concavity.

Looking at your graph of y = sin(x), this occur at x = pi.

Thus, the general solution will be:

x = n*pi, where n is any integer.