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Mathematics 29 Online
OpenStudy (anonymous):

does anyone know how to determine interval using < and > signs

OpenStudy (anonymous):

you mean solve inequalities?

OpenStudy (anonymous):

or interval notation?

OpenStudy (anonymous):

i got a question to find the missing numbers and plot the graph and determine the intervals for which f(x)\[\ge\]0

OpenStudy (anonymous):

ok, what are the numbers?

OpenStudy (anonymous):

and what is f(x)

OpenStudy (anonymous):

first you have to find when f(x)=0, this divides the domain up into intervals. you then test a sample from each interval to see if its positive, negative, or zero

OpenStudy (anonymous):

also find when f(x) is undefined

OpenStudy (anonymous):

yes i did that but it is suppose to look like this: -3x^2+4x+4<0 x<-0.666 or x>2

OpenStudy (anonymous):

x -3,-2,-1.5,-1,0,1,2 f(x) -7,0,2,3,2,-3,-12

OpenStudy (anonymous):

3x^2-4x-4>0 3x^2-4x-4=0 3x^2-6x+2x-4=0 3x(x-2)+2(x-2)=0 (x-2)(3x+2)=0 x=2, -2/3 this divides the domain up into three intervals (-infinity, -2/3), (-2/3, 2), (2, infinity)

OpenStudy (anonymous):

After plotting the graph for that it asks 3 questions: i. The coordinates of the maximum point ii. The interval for which􀀁f(x) ≥ 0 iii. The interval for which f(x) ≤ 0

OpenStudy (anonymous):

so here you test a value from each interval. for the first interval we'll test x=-1 (x-2)(3x+2)>0 (negative)(negative)>0, true, this interval is part of the solution 2nd interval, we'll use 0 (negative)(positive)>0 false, this is not part of solution 3rd interval, we'll use x=3 (positive)(positive)>0, true, this is part of solution solution set: (-infinity, -2/3)U(2, infinity)

OpenStudy (anonymous):

Im doing college algebra

OpenStudy (anonymous):

so your question was which one?

OpenStudy (anonymous):

did you plot the graph?

OpenStudy (anonymous):

the last 2, yes I did the graph already

OpenStudy (anonymous):

what was the shape of graph? parabola?

OpenStudy (anonymous):

it is mountain shaped

OpenStudy (anonymous):

let me plot it, 1 second

OpenStudy (anonymous):

f(x)=2-3x-2x^2

OpenStudy (anonymous):

yes i just did a regression and got -2x^2-3x+2, it is a parabola

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

website crashed on me

OpenStudy (anonymous):

the maximum can be solved by finding the y-coordinate of the vertex, it's a maximum because the parabola opens down

OpenStudy (anonymous):

for that i got -1,3

OpenStudy (anonymous):

you got -1, 3 for what?

OpenStudy (anonymous):

the maximum point

OpenStudy (anonymous):

not quite, from the regression we have f(x)=-2x^2-3x+2

OpenStudy (anonymous):

the vertex is at (3/[2(-2)], f(3/[2(-2)]) or (-3/4, 25/8)

OpenStudy (anonymous):

so the max value is 25/8

OpenStudy (anonymous):

ok I think you are more advance for what the instructor wants he said to look at the graph and just write down the coordinates of the max point

OpenStudy (anonymous):

ah ok, but does he want you to solve the inequalities or just look at the graph?

OpenStudy (anonymous):

yes I solved it but he want it in a format that I'm not so clear on

OpenStudy (anonymous):

interval notation?

OpenStudy (anonymous):

or set-builder notation?

OpenStudy (anonymous):

I solved it ad got: 0.5 and -2

OpenStudy (anonymous):

ok, so now you have the intervals you need to solve the inequality

OpenStudy (anonymous):

in and example he has x^2-3x+2>0 answer: x<1 or x>2

OpenStudy (anonymous):

this splits the domain into 3 intervals

OpenStudy (anonymous):

ok so you have the zeros of f(x) in your question, 1/2 and -2

OpenStudy (anonymous):

so the factored form of f(x) is (2x-1)(x+2)

OpenStudy (anonymous):

but I solved the inequality 2-3x-2x^2|dw:1332811990724:dw| and got 0.5 and -2

OpenStudy (anonymous):

you didn't solve the inequality, you solved the equation 2-3x-2x^2=0

OpenStudy (anonymous):

it was not equal to 0 it was greater than or equal to 0

OpenStudy (anonymous):

yeah, that's the next step, solving the inequality, which is easy when you found the zeros

OpenStudy (anonymous):

|dw:1332812190676:dw|

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