Question

The function f(x)=x^2 is decreasing on what interval?

Answer 1

in interval ]0;1[

Answer 2

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).

Answer 3

That's not an interval, @Alienium . If you what you mean is [0,1], that is not correct.

Answer 4

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... :/

Answer 5

but they're pretty streight forward....

Answer 6

@DebbieG and yes it is correct...

Answer 7

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.

Answer 8

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

Answer 9

my choices are
\[(-\infty,0]\]
\[(-\infty,0)\]
\[(0,\infty)\]
\[[0,\infty)\]

Answer 10

@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

Answer 11

@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??

Answer 12

@kristinaa does that make sense? What do you think the answer is?

Answer 13

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

Answer 14

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.

Answer 15

It makes since a little I am still a little confused\[[0,\infty)\]

Answer 16

is that the right answer?

Answer 17

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?

Answer 18

oh :/ it is a parabola, and it is opening up

Answer 19

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.

Answer 20

Right! And like I said, a parabola "turns around" at the vertex.

Answer 21

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?

Answer 22

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 \[-\infty\]

Answer 23

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. :)

Answer 24

yay! so I was right?!
its \[(-\infty,0)\]