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Mathematics 22 Online
OpenStudy (anonymous):

Determine the intervals on which the function is increasing, decreasing, and constant.

OpenStudy (anonymous):

OpenStudy (anonymous):

@Hero

OpenStudy (anonymous):

I don't understand how to find the answer...

OpenStudy (anonymous):

the choices are: Increasing x > 0; Decreasing x <0 Decreasing on all real numbers INcreasing on all real numbers Increasing x < 0; Decreasing x > 0

OpenStudy (anonymous):

@myininaya

OpenStudy (anonymous):

@zepdrix

OpenStudy (anonymous):

@cwrw238

OpenStudy (anonymous):

@Luigi0210

OpenStudy (anonymous):

@e.mccormick

OpenStudy (anonymous):

@Destinymasha

OpenStudy (anonymous):

Somebody please....

OpenStudy (anonymous):

@Compassionate

OpenStudy (anonymous):

Would it be only increasing? I don't understand how to read the graph and answer with x > 0 or x <0

OpenStudy (anonymous):

@ash2326

OpenStudy (ash2326):

Check the graph, with increase in x is there increase in y (or f(x))?

OpenStudy (anonymous):

I believe that there is only an increase so the y must also increase as well right? if was going to be a decrease, then the increase for x would be the decrease for y correct?

OpenStudy (anonymous):

also i don't understand when they use f(x) and f(x) = f(-x), i get confused with the number < x < number

OpenStudy (ash2326):

Ok, I will explain that doubt later

OpenStudy (anonymous):

Okay, but since the graph is only increasing then would it be increasing on all real numbers?

OpenStudy (ash2326):

Sorry

OpenStudy (anonymous):

So the answer would be increasing on all real numbers since it is only increasing?

OpenStudy (ash2326):

As you move right, the graph rises. Which means as x increases, y increases. For x=0, there are many values of y but slope of line >0, so it's increasing at x=0 as well. So it's increasing for all real numbers

OpenStudy (anonymous):

Okay, i think i understand it a bit more :)

OpenStudy (anonymous):

Thank you very much! I was going crazy not understanding the topic at all :D

OpenStudy (ash2326):

Welcome :D

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