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in interval ]0;1[
It's a quadratic, and quadratics always "turn around" at the vertex. So when you find the vertex, you know where it changes direction. Then it's just a matter of whether it goes from increasing to decreasing (if it is downward opening) or decreasing to increasing (if it is upward opening).
That's not an interval, @Alienium . If you what you mean is [0,1], that is not correct.
intervals are noted with closing and opening square brackets respectively it's because only the intermediate values are valid not the actual limits. ]0;1[ means from 0 to 1 without 0 and 1 if it was [0;1] it's a segment and it includes 0 and 1 rules of europe... :/
but they're pretty streight forward....
@DebbieG and yes it is correct...
Well as for the notation of the interval, maybe we need to know where the question asker is, because that is not how we would write "the set of all numbers between, but not including, 0 and 1" here. But regardless, that is definitely NOT where \( \Large f(x)=x^2\) is decreasing. In fact, it isn't decreasing anywhere on that interval.
we're saying the same solution from 0.0000000000000999 or sth to 0.99999 it is decrasing because 0.5^2 = 0.25 so it means the values are getting lower it's about the trend of the function not about the value
my choices are \[(-\infty,0]\] \[(-\infty,0)\] \[(0,\infty)\] \[[0,\infty)\]
@kristinaa , think about what a parabola looks like. If opens downward, then it is: increasing on -infinity to x-coordinate of the vertex decreasing on x-coordinate of the vertex to infinity If opens upward then it is: decreasing on -infinity to x-coordinate of the vertex increasing on x-coordinate of the vertex to infinity
@Alienium I have no idea what you mean. "from 0.0000000000000999 or sth to 0.99999 it is decrasing because 0.5^2 = 0.25 so it means the values are getting lower" f(0)=0 f(.5)=0.25 f(.8)=0.64 f(1)=1 Are those values decreasing??
@kristinaa does that make sense? What do you think the answer is?
yes... you're not getting the point ok... you tell me how you conceptualize the problem and i'll tell you what i mean maybe you're right... just tell me how you're thinking about it
It's now "how I'm thinking about it"... it's simply what the question is asking. A function is INCREASING on an interval if, as the the x values increase, the y values increase. E.g., in a LINEAR function, it means positive slope. In a non-linear function it means positive 1st derivative, but you don't need to get into that for a simple quadratic... just look at what the y's do as you move left to right along the x-axis. A function is DECREASING on an interval if, as the the x values increase, the y values decrease. E.g., in a LINEAR function, it means negative slope (or negative 1st derivative). This isn't my opinion, google it! It's what it MEANS.
It makes since a little I am still a little confused\[[0,\infty)\]
is that the right answer?
No, that isn't it. do you know what the shape of y=x^2 is? What does it look like? open up or open down?
oh :/ it is a parabola, and it is opening up
I always tell my students, "read a function like you read a book - from left to right". So "increasing" means that the graph is "going up" as you look at it left to right, and "decreasing" means that the graph is "going down" as you look at it left to right.
Right! And like I said, a parabola "turns around" at the vertex.
So see what I wrote above: If opens upward then it is: decreasing on -infinity to x-coordinate of the vertex increasing on x-coordinate of the vertex to infinity Now you know this one opens up.... so.....? Where is it decreasing?
does that mean that it would be \[(-\infty,0)\] because I read above that it is decreasing so wouldn't that mean that the first value is \[-\infty\]
Exactly! And your choices give you that same interval, one with a ,0) and one with a ,0]. The bracket would mean that the 0 is INCLUDED in the interval. But the problem with that is that the function "turns around" right AT x=0, since that's the vertex. So we don't say that it's "increasing" OR "decreasing" there - it's neither. Because really, it's decreasing as we are coming at 0 from the negative side of the x-axis, but it's increasing as we move away from 0 toward the positive side of the x-axis. So that's why you want the ,0) option and not the ,0] option. :)
yay! so I was right?! its \[(-\infty,0)\]