heyitslizzy13
  • heyitslizzy13
can someone help me with a problem please!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
heyitslizzy13
  • heyitslizzy13
heyitslizzy13
  • heyitslizzy13
@satellite73 @Directrix @jim_thompson5910
Jhannybean
  • Jhannybean
which part of the problem?

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heyitslizzy13
  • heyitslizzy13
The first part for now :)
freckles
  • freckles
Hint: Intermediate Value theorem
heyitslizzy13
  • heyitslizzy13
@freckles do you know any good websites to review the theorem?
Jhannybean
  • Jhannybean
\[f(x) = ax^3-11x^2+bx+c \qquad f(-3)=10, \qquad f(0)=4,\qquad f(-1)=4\]\[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10 \]\[f(-3)=-27a-99-3b+c=10\]\[f(-3) = -27a-3b+c=109\] ...perhaps we can solve this as a system of equations?
heyitslizzy13
  • heyitslizzy13
could you show me how? please and thank you :)
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
now let's say we had point A and point B one point on either side |dw:1443850712927:dw|
jim_thompson5910
  • jim_thompson5910
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
heyitslizzy13
  • heyitslizzy13
okay
jim_thompson5910
  • jim_thompson5910
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
Jhannybean
  • Jhannybean
\[f(0) = a(0)^3 -11(0)+b(0)+c =4\]\[f(0) = c=4\] \[f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4\]\[f(-1) = -a-11-b+c=4\]\[f(-1) =-a-b+c=15\] \[\therefore \]\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\] 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.
heyitslizzy13
  • heyitslizzy13
I am kind of confused, how do we know that c=4 is the root that lies in the middle?
Jhannybean
  • Jhannybean
because when \(x=0\) (that means it's at the origin) \(y=4\)
heyitslizzy13
  • heyitslizzy13
ohhh okay
jim_thompson5910
  • jim_thompson5910
Jhannybean plugged in x = 0 and everything went away basically except for that last term
Jhannybean
  • Jhannybean
Exactly
heyitslizzy13
  • heyitslizzy13
do you understand what part a of the problem is asking? I'm quite confused
Jhannybean
  • Jhannybean
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.
jim_thompson5910
  • jim_thompson5910
if you need a visual way to approach this, draw a blank xy axis |dw:1443851396111:dw|
heyitslizzy13
  • heyitslizzy13
at -1 i believe , (-1, 4) ?
jim_thompson5910
  • jim_thompson5910
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|
jim_thompson5910
  • jim_thompson5910
now, ask yourself this: is it possible to connect the two points without crossing the x axis?
heyitslizzy13
  • heyitslizzy13
it is possible
jim_thompson5910
  • jim_thompson5910
so there is no guarantee we have a root on the interval from x = -3 to x = 0
jim_thompson5910
  • jim_thompson5910
now IF we had (0,-4) instead, then we shift down the second point |dw:1443851620081:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1443851638079:dw|
jim_thompson5910
  • jim_thompson5910
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|
heyitslizzy13
  • heyitslizzy13
ohhh okay!
jim_thompson5910
  • jim_thompson5910
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)
jim_thompson5910
  • jim_thompson5910
sign change in the y values or function outputs I mean
jim_thompson5910
  • jim_thompson5910
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
heyitslizzy13
  • heyitslizzy13
oh i see now
heyitslizzy13
  • heyitslizzy13
could you help me on part b?
jim_thompson5910
  • jim_thompson5910
ok so I think you found that c = 4 right?
jim_thompson5910
  • jim_thompson5910
do you see how Jhannybean got this? \[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\]
heyitslizzy13
  • heyitslizzy13
yes :)
heyitslizzy13
  • heyitslizzy13
wait i don't understand how she got the 109, 4 and15
jim_thompson5910
  • jim_thompson5910
I think f(-3) should have been 10 and not 109
heyitslizzy13
  • heyitslizzy13
okay, and how is f(-1)= 15?
jim_thompson5910
  • jim_thompson5910
it should be \[f(-3) = -27a-3b+c=10\\ f(0) =c=4 \\f(-1)=-a-b+c=4\]
heyitslizzy13
  • heyitslizzy13
oh okay, i understand then
jim_thompson5910
  • jim_thompson5910
oh wait, there's an 11 in there that I missed, let me rethink
jim_thompson5910
  • jim_thompson5910
Jhannybean has it right, I missed a term
Jhannybean
  • Jhannybean
\[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10\]\[f(-3) = -27a -99 -3b +c = 10\]\[f(-3) = -27a -99 -3b + c +99 = 10 + 99\]\[f(-3) = -27a -3b+c=109\]
Jhannybean
  • Jhannybean
\(\color{#0cbb34}{\text{Originally Posted by}}\) @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. \(\color{#0cbb34}{\text{End of Quote}}\)
heyitslizzy13
  • heyitslizzy13
oh ok
Jhannybean
  • Jhannybean
Does that make more sense, @heyitslizzy13 ?
heyitslizzy13
  • heyitslizzy13
yes :)
Jhannybean
  • Jhannybean
Great \(\checkmark\)
jim_thompson5910
  • jim_thompson5910
`that root would be c=4 ` I don't know about that part
heyitslizzy13
  • heyitslizzy13
so we know c is 4, what about a and b?
Jhannybean
  • Jhannybean
Haha, I meant \(x=0\) x_x
jim_thompson5910
  • jim_thompson5910
ah I gotcha
heyitslizzy13
  • heyitslizzy13
would it be 109 and 15?
jim_thompson5910
  • jim_thompson5910
since c = 4 we can replace each 'c' with 4 \[-27a-3b+\color{red}{c}=109\\-a-b+\color{red}{c}=15\] \[-27a-3b+\color{red}{4}=109\\-a-b+\color{red}{4}=15\]
Jhannybean
  • Jhannybean
So then you would solve it either by substitution, or using the elimination method :)
heyitslizzy13
  • heyitslizzy13
can you walk me through either one please? I know what they are but I'm a little lost rn
Jhannybean
  • Jhannybean
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?
heyitslizzy13
  • heyitslizzy13
yes!
Jhannybean
  • Jhannybean
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 \(a\).
Jhannybean
  • Jhannybean
Is this making sense, @heyitslizzy13 ?
heyitslizzy13
  • heyitslizzy13
yes!! :)) is a=-3?
Jhannybean
  • Jhannybean
Yes it is :)
Jhannybean
  • Jhannybean
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 \(-a-b=11\) . \[-(-3)-b=11\]\[3-b=11\]\[-3+3-b=11-3\]\[-b=8\]\[b=-8\]
heyitslizzy13
  • heyitslizzy13
okay thanks :)))
heyitslizzy13
  • heyitslizzy13
can you help me on c also?
jim_thompson5910
  • jim_thompson5910
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
heyitslizzy13
  • heyitslizzy13
thanks!!
jim_thompson5910
  • jim_thompson5910
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
heyitslizzy13
  • heyitslizzy13
thank you both so much!!!! :)))

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