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 2 years ago
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
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).

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

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

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0let's see\[f(12)=102\le182\le202\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I think I need to stop multitasking to do this

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0\[227 ≤ f(17) ≤ 462\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I have a strong feeling that @asnaseer is going to save us

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

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

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

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0That would make f(2) = 2?

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0When you plug it back in anyway...

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3and 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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3we have two equations so far

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3maybe make use of f(0)?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3which gives us:\[6\le c\le 10\]

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

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0I 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?

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3what do you mean by /strange/?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3is there a particular approach that is used?

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3hmmm... let me think over that approach...

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3I 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:\[a+b+c=4\tag{3b}\]

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0Does it become a geometric question when they complete the square? dw:1343584198825:dw

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0They 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.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0just thinking out loud, thought they might subtract squares then.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0hmm no, that's not what I thought then.

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0I don't unserstand how they MUST have the same vertex though...

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3interesting observation @slaaibak  maybe we can use that to get the 3rd equation?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3yes @Calcmathlete that ties up with @slaaibak's observation

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1if 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

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.0ya so they all have same vertex. as @Calcmathlete 's.............. said:D

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3the LHS and RHS have the same vertex which fixes the vertex of the middle one

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

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.0thus they all have same vertex

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1both 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

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1343584535907:dw

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.03 functions quadratic......

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

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0Alright. 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?

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1f(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

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.0Alright. Thank you! I wish I could give multiple medals...

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

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

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.0Graphs must be there.
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