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Calcmathlete
Group Title
Let f(x) be a quadratic function. If \(x^2  4x + 6 ≤ f(x) ≤ 2x^2  8x +10\) for all real numbers x, and f(12) = 182, find f(17).
 2 years ago
 2 years ago
Calcmathlete Group Title
Let f(x) be a quadratic function. If \(x^2  4x + 6 ≤ f(x) ≤ 2x^2  8x +10\) for all real numbers x, and f(12) = 182, find f(17).
 2 years ago
 2 years ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.0
my first thought is to try factoring and see if that takes us somewhere
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
Aren't they both unfactorable?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I guess, I'm just brainstorming (and multitasking)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
changing the second into 2(x^24x+5) may shed some light is thought #2
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
let's see\[f(12)=102\le182\le202\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I think I need to stop multitasking to do this
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
\[227 ≤ f(17) ≤ 462\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I have a strong feeling that @asnaseer is going to save us
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
could we make use of the fact that if the LHS=RHS then that fixes f(x)?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
i.e.\[x^24x+6=2x^28x+10\]f we solve this we get:\[(x2)^2=0\implies x=2\]
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
so at x=2 both the LHS (left hand side) and RHS (right hand side) are equal
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
interesting...
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
That would make f(2) = 2?
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
When you plug it back in anyway...
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
and if we define f(x) as:\[f(x)=ax^2+bx+c\]then maybe we can form 3 equations and solve for a, b and c
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
we have two equations so far
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
maybe make use of f(0)?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
which gives us:\[6\le c\le 10\]
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
\[x^2  4x + 6 ≤ (ax^2+bx+c)≤ 2x^2  8x +10\] \[\color{blue}{a(12)^2+b(12)+c}=182=\color{blue}{144a+12b+c.}\] \[\color{red}{a(17)^2+b(17)+c=?}\]\[6 \le c \le 10.\]
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
I just looked up how they did it...it's very strange, and I'm having a bit of trouble understanding a part of it...do you want me to post it or would you like to try figuring the rest out?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
f(1) is also interesting, it gives us:\[3\le a+b+c\le4\]
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
what do you mean by /strange/?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
is there a particular approach that is used?
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
Well, the way they did it is a lot simpler and way different from this way...they completed the square...
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
hmmm... let me think over that approach...
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Becoz we all are doing this type of question first time. & they had done it already. I really appreciate @asnaseer 's work:D
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
all three equations have a turning point (minimum) at x=2. so you could use that.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
I can't /see/ how completing the square helps. but I wonder if we can make use of my last result for f(1) to get 3 possible equations to solve.\[144a+12b+c=182\tag{1}\]\[4a+2b+c=2\tag{2}\]and then either:\[a+b+c=3\tag{3a}\]or:\[a+b+c=4\tag{3b}\]
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
Does it become a geometric question when they complete the square? dw:1343584198825:dw
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
They said that once you complete the square, you find that f(x) must have the same vertex as the other two equations and that you just have to solve for a.
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
just thinking out loud, thought they might subtract squares then.
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
hmm no, that's not what I thought then.
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
I don't unserstand how they MUST have the same vertex though...
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
interesting observation @slaaibak  maybe we can use that to get the 3rd equation?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
yes @Calcmathlete that ties up with @slaaibak's observation
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
if we use f(x) = a(xb)^2 + k, we can solve. because we have b:2 and k can be found by f(2) = k so it's pretty easy from here
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
ya so they all have same vertex. as @Calcmathlete 's.............. said:D
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
Thin is, I don't understand how they MUST all have the same vertex. Is it because of how the parabolas work when graphed?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
the LHS and RHS have the same vertex which fixes the vertex of the middle one
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
no becoz they all have same minimum value at x=2
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
thus they all have same vertex
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
both have the same turning point. therefore, f(x) can't be decreasing if the other two functions are increasing, because f(2) are the same for all the equations. therefore all of them have a minimum at 2
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
dw:1343584535907:dw
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
3 parabolas
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
3 functions quadratic......
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
very similar to the fact that x=2 gives the same value for LHS and RHS which fixes the value of f(2)
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
Alright. That's really why i didn't understand. I guess since the opposite ends of the compound inequality have the same vertex, it is justified. From there it is easy I guess. \[f(12) = a(x  2)^2 + 2 \implies a = 1.8\]Then we can just plug in 17? \[f(17) = 1.8(225) + 2 = 405 Is that right?
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
f(2) = 2 f(x) = a(x2)^2 + k b=2 due to minimum so f(2) = a(22)^2 + k = 2 therefore k = 2 now f(12) = 182 so 182 = a(122)^2 + 2 182 = 100a + 2 100a = 180 a = 1.8
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
yes @Calcmathlete
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
yeah that's correct
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
Alright. Thank you! I wish I could give multiple medals...
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
yw :) and yes, I wish we had Olympicstyle medals  Gold/Silver/Bronze :)
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
but then we would have to drug test everyone to make sure they were not cheating! :D
 one year ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
lol...
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Graphs must be there.
 one year ago
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