can someone help me with a problem please!!

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can someone help me with a problem please!!

Mathematics
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which part of the problem?

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The first part for now :)
Hint: Intermediate Value theorem
@freckles do you know any good websites to review the theorem?
\[f(x) = ax^3-11x^2+bx+c \qquad f(-3)=10, \qquad f(0)=4,\qquad f(-1)=4\]\[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10 \]\[f(-3)=-27a-99-3b+c=10\]\[f(-3) = -27a-3b+c=109\] ...perhaps we can solve this as a system of equations?
could you show me how? please and thank you :)
part a) imagine you have a river that is a complete straight line |dw:1443850668196:dw| this straight line would stretch forever in both directions
now let's say we had point A and point B one point on either side |dw:1443850712927:dw|
if 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
okay
now 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
\[f(0) = a(0)^3 -11(0)+b(0)+c =4\]\[f(0) = c=4\] \[f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4\]\[f(-1) = -a-11-b+c=4\]\[f(-1) =-a-b+c=15\] \[\therefore \]\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\] 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.
I am kind of confused, how do we know that c=4 is the root that lies in the middle?
because when \(x=0\) (that means it's at the origin) \(y=4\)
ohhh okay
Jhannybean plugged in x = 0 and everything went away basically except for that last term
Exactly
do you understand what part a of the problem is asking? I'm quite confused
Okay so between \(x=-3\) and \(x=0\), is there ANYWHERE on the x-axis when the function gives you a solid number? hint: look at the above posts and our explanations.
if you need a visual way to approach this, draw a blank xy axis |dw:1443851396111:dw|
at -1 i believe , (-1, 4) ?
then plot the point (-3,10), from the fact that f(-3) = 10 and the point (0,4), because f(0) = 4 |dw:1443851450292:dw|
now, ask yourself this: is it possible to connect the two points without crossing the x axis?
it is possible
so there is no guarantee we have a root on the interval from x = -3 to x = 0
now IF we had (0,-4) instead, then we shift down the second point |dw:1443851620081:dw|
|dw:1443851638079:dw|
now 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|
ohhh okay!
so 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)
sign change in the y values or function outputs I mean
but 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
oh i see now
could you help me on part b?
ok so I think you found that c = 4 right?
do you see how Jhannybean got this? \[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\]
yes :)
wait i don't understand how she got the 109, 4 and15
I think f(-3) should have been 10 and not 109
okay, and how is f(-1)= 15?
it should be \[f(-3) = -27a-3b+c=10\\ f(0) =c=4 \\f(-1)=-a-b+c=4\]
oh okay, i understand then
oh wait, there's an 11 in there that I missed, let me rethink
Jhannybean has it right, I missed a term
\[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10\]\[f(-3) = -27a -99 -3b +c = 10\]\[f(-3) = -27a -99 -3b + c +99 = 10 + 99\]\[f(-3) = -27a -3b+c=109\]
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean \[f(0) = a(0)^3 -11(0)+b(0)+c =4\]\[f(0) = 0+0+0+c=4 \\ f(0)=c=4\] \[f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4\]\[f(-1) = -a-11-b+c+11=4+11\]\[f(-1) =-a-b+c=15\] \[\therefore\]\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\] 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. \(\color{#0cbb34}{\text{End of Quote}}\)
oh ok
Does that make more sense, @heyitslizzy13 ?
yes :)
Great \(\checkmark\)
`that root would be c=4 ` I don't know about that part
so we know c is 4, what about a and b?
Haha, I meant \(x=0\) x_x
ah I gotcha
would it be 109 and 15?
since c = 4 we can replace each 'c' with 4 \[-27a-3b+\color{red}{c}=109\\-a-b+\color{red}{c}=15\] \[-27a-3b+\color{red}{4}=109\\-a-b+\color{red}{4}=15\]
So then you would solve it either by substitution, or using the elimination method :)
can you walk me through either one please? I know what they are but I'm a little lost rn
First 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. \[-27a-3b+4-4=109 -4\]\[-a-b+4-4=15-4\] Simplifying both of these, we get: \[-27a-3b=105\]\[-a-b=11\] With me so far?
yes!
So 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 \(-3\). \[-27a-3b=105\]\[\color{red}{-3}(-a-b=11)\] \[-27a-3b=105\]\[\color{red}{+3}a\color{red}{+3}b=\color{red}{-33}\] Adding these two equations together, we will get \[(-27+3)a +(-3+3)b =105-33\]\[-24a+0b=72\]Now we just solve for \(a\).
Is this making sense, @heyitslizzy13 ?
yes!! :)) is a=-3?
Yes it is :)
Now 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 \(-a-b=11\) . \[-(-3)-b=11\]\[3-b=11\]\[-3+3-b=11-3\]\[-b=8\]\[b=-8\]
okay thanks :)))
can you help me on c also?
for 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
thanks!!
the 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
thank you both so much!!!! :)))

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