Board 8 > quick calc question that I'm drawing a blank on

(+) Yeah! (+)

this one's easy.. i'm not sure what to do though I know I should:

find the interval in which the function f(x) = | x +1| is increasing

how do i do this again? hints?

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f(x) increases on the interval (-1, infinity), just for clarity's sake

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~Peaf~

and (-infinity, -1)

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*Lives in the Future*

find when df/dx is > 0



since it's absolute value, it might be easier to create two separate functions

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~Peaf~

it is not increasing on that interval, it is decreasing

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~Peaf~

I'm assuming that abs(x+1)

Which absolute value of -infinity is infinity.

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*Lives in the Future*

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You're messing with me! You're messing with me, aren't you!?
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Wait, for some reason I was thinking of starting at -1 and going towards -infinity

I really need some sleep

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*Lives in the Future*

[This message was deleted at the request of the original poster]

wait i'm still confused

can someone go through this step by step?

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f(x) = | x + 1|

Define f(x) piecewise.

For x < -1, f(x) = -x - 1
For x >= -1, f(x) = x + 1

Take derivatives.

For x < -1, f(x) = -1
For x > -1, f(x) = 1

Derivative is positive for x > -1

The function isn't differentiable at x = -1 so you can't really say it is increasing or decreasing there.

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No amount of planning will ever replace dumb luck.

sure

ok, separate f(x) = |x+1| into two functions

first is f(x) = x+1

the other is f(x) = -(x+1)

find where they intersect so you know where one stops and the other begins
x+1 = -x - 1
2x = -2
x=-1

and we know that it's essentially linear, so each function is real from (-infinity, infinity)

however, since they intersect at x=-1, we have a point discontinuity at x=-1 (there's a way to prove this, I forget exactly how, but if you just look at the graph you'll see it). There exists no derivative for f(x) at a discontinuity.

so the interval for dy/dx (if we let y=f(x) is (-infinity, -1), (-1, infinity). Now we want to find where f(x) increases, so find the derivative of each function from earlier

dy/dx = 1
dy/dx = -1

f(x) increases when its derivative is positive. 1>0 for all real values, -1 is never >0. So only the first function (f(x) = x+1) increases. |x+1| = x+1 on the interval from (-1, infinity).

f(x) increases on the interval (-1, infinity)

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~Peaf~

PIECEWISE, THAT'S THE WORD I COULDN'T REMEMBER

**** me

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~Peaf~

so in interval notation.. the answer would be

(-1, infinity]

?

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also, I think I meant "corner discontinuity", not "point discontinuity"

'meh'

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~Peaf~

no

you can't have a square bracket around infinity, ever

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~Peaf~

From: HeroicLinusReed | #014
so in interval notation.. the answer would be

(-1, infinity]

?

Putting a closed bracket on infinity isn't good, since it implies there's an exact upper bound... which there isn't since it is infinite. Always use open for infinity. Otherwise, yeah, it's good.

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No amount of planning will ever replace dumb luck.

ooooooh ok sorry i was typing it before i saw your post, frog. that post explains it very well.

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and yeah, Ness is a bit more eloquent with his description of the bracket


yeah, glad we can help. I know a lot of people here are math/engineer majors. b8chat saved my ass in physics last year...

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~Peaf~

Guess you guys have already got this, ha.

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last.fm/user/BlackHydras

one more question:

if they give me the picture of a parabola, how am i supposed to figure out the quadratic equation for this parabola? what are the steps to figfuring this out?

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umm...

do they give you multiple pictures and ask you to match them graph with the equation? because I remember having a lot of problems like that

if it's just a graph, your best bet is to just write specific points in an x,y table and try to figure out the pattern

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~Peaf~

If you can see the maximum or minimum point of the parabola, let that point be (x0,y0). Then the equation will be of the form

y = m(x-x0)^2+y0

Then use another point to find m. If the parabola is open upwards (like a smile), m will be positive; if the parabola is open downwards (like a frown), m will be negative.

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Worry is a misuse of imagination.
Fantasizing about Guru champion BlAcK TuRtLe, however, is a great use of it.

Pretty sure with infinity, square or round bracket doesn't matter since you can't ever have a solid bound on infinity so it all means the same thing

From: HeroicLinusReed | #021
if they give me the picture of a parabola, how am i supposed to figure out the quadratic equation for this parabola? what are the steps to figfuring this out?

Is it a to-scale graph? Pick 3 points and plug them into the standard quadratic equation, solve the linear system for a,b,c

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_foolmo_
'so I can potentially win the guru without getting the final right?' - BlAcK TuRtLe