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Oct 6 2014 06:35am
hey friends... im supposed to graph a function but i don't know how to do it. there are three different equations but it's apparently the same function? the question is:

graph the function f(x) = 1+x if x is less than or equal to negative one, f(x) = x^2 if x is greater tha or equal to -1 but less than 1, and f(x) = 2-x if x is greater than or equal to 1.

i made a table of values for all of those equations and got different answers, and i dont know how im supposed to graph them all in one graph. if i graph them all separately but put them on the same graph (3 different graphs on the same grid) i think i would get it wrong. then i have to check the continuity of the function at x = -1 and x = 1.

(note, the question is written with a single "f(x) =" with the other parts of the equations on lines separate from each other. it looks kind of like this:

Graph the function f(x) = { 1+x if x <= -1
x^2 if -1 <= x < 1
2-x if x>= 1


except everything is in the bracket.
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Oct 6 2014 09:31am
don't look too indepth if the directions doesn't say so.
just graph it , discontinuites are legitmate
it is stilla function since no two sections overlap. it just might be be continious

This post was edited by saber_x3 on Oct 6 2014 09:32am
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Oct 6 2014 03:25pm
You're graphing three separate line segments basically.
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Oct 6 2014 05:13pm
Quote (Mastersam93 @ Oct 6 2014 05:25pm)
You're graphing three separate line segments basically.


Quote (saber_x3 @ Oct 6 2014 11:31am)
don't look too indepth if the directions doesn't say so.
just graph it , discontinuites are legitmate
it is stilla function since no two sections overlap. it just might be be continious


thanks to you guys i managed to get that part done and i feel confident about it. now im confused about something that seems like more of an algebra problem -- using P = a + bY + cY^2, i have to try to find the values of a, b, and c. P is the population and Y is the year, if that helps. I dont know where to start -- do i manipulate the equation so that the variables im solving for are alone? like a = xxxxxx and b = xxxxxxx and c = xxxx?
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Oct 6 2014 06:51pm
Quote (Swxg @ Oct 6 2014 07:13pm)
thanks to you guys i managed to get that part done and i feel confident about it. now im confused about something that seems like more of an algebra problem -- using P = a + bY + cY^2, i have to try to find the values of a, b, and c. P is the population and Y is the year, if that helps. I dont know where to start -- do i manipulate the equation so that the variables im solving for are alone? like a = xxxxxx and b = xxxxxxx and c = xxxx?


P(Y) is a quadratic. if you need to determine the coefficients, you'll need some data. what data do you have?
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Oct 6 2014 07:13pm
Quote (Swxg @ Oct 6 2014 09:05pm)
for the calc homework, i have a table of years and population, starting in the year 1900 what do i do with the data?


you can make it a linear equation and solve it.

eg:


P(Y) = a + bY + cY^2

suppose these were your values:
Y, P(Y)
1, 3
2, 10
3, 27

now you have:
3 = a + b + c
10 = a + 2b + 4c
27 = a + 3b + 16c

now solve using your favourite method (row reduce, elimination, substitution)
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Oct 6 2014 07:23pm
Quote (carteblanche @ Oct 6 2014 09:13pm)
you can make it a linear equation and solve it.

eg:


P(Y) = a + bY + cY^2

suppose these were your values:
Y, P(Y)
1, 3
2, 10
3, 27

now you have:
3 = a + b + c
10 = a + 2b + 4c
27 = a + 3b + 16c

now solve using your favourite method (row reduce, elimination, substitution)


ok thanks -- now, one moreq eustion.
t2 ∝ d3 or t ∝ d3/2. Use the model t = adb to find the values of a and b. check if b is nearly equal to 3/2 or not.
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Oct 6 2014 09:33pm
Quote (Swxg @ Oct 6 2014 09:23pm)
ok thanks -- now, one moreq eustion.
t2 ∝ d3 or t ∝ d3/2. Use the model t = adb to find the values of a and b. check if b is nearly equal to 3/2 or not.


Quote (Swxg @ Oct 6 2014 09:23pm)
ok thanks -- now, one moreq eustion.
t2 ∝ d3 or t ∝ d3/2. Use the model t = adb to find the values of a and b. check if b is nearly equal to 3/2 or not.


bump ast part
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Oct 7 2014 05:32am
HALP
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Oct 7 2014 11:15am
logan just ask me dood
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