Question

locate the critical points f(x)=lsinxl

Answer 1

if you understand what an absolute value does, then you'll realize that it has no real importance in the problem and you can ignore it

Answer 2

so do i just find the first derivative of sinx?

Answer 3

its basically only the maximums for sin(x) as the curve doesn't go below the x-axis.

at the roots points of inflexion occur.

Answer 4

points of inflection can also be determined by just setting f(x)= 0
2nd derivative is just negative sine

Answer 5

@campbell_st you want the minimums too, since the minimums are also now maximums

@completeidiot There should be no points of inflection, |sinx| should always be concave up.

Answer 6

okay so how do i solve this?

Answer 7

*concave down.

Answer 8

Just differentiate and find where f' = 0. You can basically ignore the abs value.

https://www.google.com/search?q=abs(sin(x))&aq=f&oq=abs(sin(x))&aqs=chrome.0.57j0l3j62l2.3500j0&sourceid=chrome&ie=UTF-8

All the peaks are where f' = 0

Answer 9

f(x)=sinx

Differentiate that and set f'(x) = 0 - the critical points will be the same x values as the absolute value function.

Answer 10

There aren't any points of inflection, since a point of inflection is where the graph goes from concave up to concave down - abs(sin(x)) is always concave down: http://www.wolframalpha.com/input/?i=abs%28sin%28x%29%29+points+of+inflection

Answer 11

We seem to be missing the Critical Points the derivative doesn't know about.

\(x = 0 + k\pi\), where \(k is an integer.\)

Answer 12

Well yeah but I left that for her to figure out :P

Answer 13

All the points where the function is not differentiable occur when f(x) = 0, ie when sinx = 0.