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heyitslizzy13
 one year ago
can someone help me with a problem please!!
heyitslizzy13
 one year ago
can someone help me with a problem please!!

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heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 @Directrix @jim_thompson5910

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which part of the problem?

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0The first part for now :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0Hint: Intermediate Value theorem

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0@freckles do you know any good websites to review the theorem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x) = ax^311x^2+bx+c \qquad f(3)=10, \qquad f(0)=4,\qquad f(1)=4\]\[f(3)=(3)^3a 11(3)^2+(3)b+c=10 \]\[f(3)=27a993b+c=10\]\[f(3) = 27a3b+c=109\] ...perhaps we can solve this as a system of equations?

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0could you show me how? please and thank you :)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3part a) imagine you have a river that is a complete straight line dw:1443850668196:dw this straight line would stretch forever in both directions

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3now let's say we had point A and point B one point on either side dw:1443850712927:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3if we connect the two points with a continuous function, then it is impossible to not pass through the line. we might have something like this dw:1443850761686:dw the path has to cross over the river somewhere

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3now imagine the river is the x axis if point A has a positive y coordinate (ie it's above the x axis) and point B has a negative y coordinate (ie it's below the x axis) then the continuous function must cross the x axis at some point between A and B. So if we know A is above the x axis and B is below, there is at least one root in between

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(0) = a(0)^3 11(0)+b(0)+c =4\]\[f(0) = c=4\] \[f(1) = a(1)^3 11(1)^2+b(1)+c=4\]\[f(1) = a11b+c=4\]\[f(1) =ab+c=15\] \[\therefore \]\[f(3) = 27a3b+c=109 \\ f(0) =c=4 \\f(1)=ab+c=15\] As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be \(c=4\) if i'm not mistaken.

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0I am kind of confused, how do we know that c=4 is the root that lies in the middle?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because when \(x=0\) (that means it's at the origin) \(y=4\)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Jhannybean plugged in x = 0 and everything went away basically except for that last term

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0do you understand what part a of the problem is asking? I'm quite confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so between \(x=3\) and \(x=0\), is there ANYWHERE on the xaxis when the function gives you a solid number? hint: look at the above posts and our explanations.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3if you need a visual way to approach this, draw a blank xy axis dw:1443851396111:dw

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0at 1 i believe , (1, 4) ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3then plot the point (3,10), from the fact that f(3) = 10 and the point (0,4), because f(0) = 4 dw:1443851450292:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3now, ask yourself this: is it possible to connect the two points without crossing the x axis?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so there is no guarantee we have a root on the interval from x = 3 to x = 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3now IF we had (0,4) instead, then we shift down the second point dw:1443851620081:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443851638079:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3now it's impossible to connect the two points without crossing the x axis so there has to be at least one root between the two x values dw:1443851682979:dw or dw:1443851692524:dw or dw:1443851699567:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so really the only guarantee that we have a root is if there is a sign change (in my example the sign change goes from y = 10 to y = 4)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3sign change in the y values or function outputs I mean

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3but since y = 10 to y = 4 has no sign changes in the problem, there is no guarantee there's a root between x = 3 to x = 0

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0could you help me on part b?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3ok so I think you found that c = 4 right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3do you see how Jhannybean got this? \[f(3) = 27a3b+c=109 \\ f(0) =c=4 \\f(1)=ab+c=15\]

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0wait i don't understand how she got the 109, 4 and15

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3I think f(3) should have been 10 and not 109

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0okay, and how is f(1)= 15?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3it should be \[f(3) = 27a3b+c=10\\ f(0) =c=4 \\f(1)=ab+c=4\]

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0oh okay, i understand then

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3oh wait, there's an 11 in there that I missed, let me rethink

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Jhannybean has it right, I missed a term

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(3)=(3)^3a 11(3)^2+(3)b+c=10\]\[f(3) = 27a 99 3b +c = 10\]\[f(3) = 27a 99 3b + c +99 = 10 + 99\]\[f(3) = 27a 3b+c=109\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean \[f(0) = a(0)^3 11(0)+b(0)+c =4\]\[f(0) = 0+0+0+c=4 \\ f(0)=c=4\] \[f(1) = a(1)^3 11(1)^2+b(1)+c=4\]\[f(1) = a11b+c+11=4+11\]\[f(1) =ab+c=15\] \[\therefore\]\[f(3) = 27a3b+c=109 \\ f(0) =c=4 \\f(1)=ab+c=15\] As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be \(c=4\) if i'm not mistaken. \(\color{#0cbb34}{\text{End of Quote}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does that make more sense, @heyitslizzy13 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great \(\checkmark\)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3`that root would be c=4 ` I don't know about that part

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0so we know c is 4, what about a and b?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, I meant \(x=0\) x_x

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0would it be 109 and 15?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3since c = 4 we can replace each 'c' with 4 \[27a3b+\color{red}{c}=109\\ab+\color{red}{c}=15\] \[27a3b+\color{red}{4}=109\\ab+\color{red}{4}=15\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then you would solve it either by substitution, or using the elimination method :)

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0can you walk me through either one please? I know what they are but I'm a little lost rn

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First we want to isolate a and b in `both` equations to one side, therefore we need to start by subtracting 4 to both sides of both the equations. \[27a3b+44=109 4\]\[ab+44=154\] Simplifying both of these, we get: \[27a3b=105\]\[ab=11\] With me so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I personally like using the elimination method. How that works is we're trying to eliminate one of the variables by using a multiplier so when we add the two equations together, we'll be left with one variable, then we can use that to solve for the other. Here, i'm going to eliminate \(b\). in order to eliminate \(b\), i need to multiply the second equation by \(3\). \[27a3b=105\]\[\color{red}{3}(ab=11)\] \[27a3b=105\]\[\color{red}{+3}a\color{red}{+3}b=\color{red}{33}\] Adding these two equations together, we will get \[(27+3)a +(3+3)b =10533\]\[24a+0b=72\]Now we just solve for \(a\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is this making sense, @heyitslizzy13 ?

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0yes!! :)) is a=3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we just plug \(a=3\) back into one of the two equations and we can solve for \(b\) I'm going to plug it into \(ab=11\) . \[(3)b=11\]\[3b=11\]\[3+3b=113\]\[b=8\]\[b=8\]

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks :)))

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0can you help me on c also?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3for part c) I would graph the function with a graphing calculator if you don't have one, use something like desmos https://www.desmos.com/calculator

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3the cool thing about desmos, as compared to a TI83, is that you can use sliders to change the graph https://www.desmos.com/calculator/tvrsgalven

heyitslizzy13
 one year ago
Best ResponseYou've already chosen the best response.0thank you both so much!!!! :)))
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