Question

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).

Answer 1

my first thought is to try factoring and see if that takes us somewhere

Answer 2

Aren't they both unfactorable?

Answer 3

I guess, I'm just brainstorming (and multitasking)

Answer 4

changing the second into
2(x^2-4x+5) may shed some light is thought #2

Answer 5

let's see\[f(12)=102\le182\le202\]

Answer 6

hm...

Answer 7

I think I need to stop multitasking to do this

Answer 8

\[227 ≤ f(17) ≤ 462\]

Answer 9

I have a strong feeling that @asnaseer is going to save us

Answer 10

could we make use of the fact that if the LHS=RHS then that fixes f(x)?

Answer 11

i.e.\[x^2-4x+6=2x^2-8x+10\]f we solve this we get:\[(x-2)^2=0\implies x=2\]

Answer 12

so at x=2 both the LHS (left hand side) and RHS (right hand side) are equal

Answer 13

interesting...

Answer 14

That would make f(2) = 2?

Answer 15

When you plug it back in anyway...

Answer 16

yes

Answer 17

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

Answer 18

we have two equations so far

Answer 19

maybe make use of f(0)?

Answer 20

which gives us:\[6\le c\le 10\]

Answer 21

\[x^2 - 4x + 6 ≤ (ax^2+bx+c)≤ 2x^2 - 8x +10\]

\[\color{blue}{a(12)^2+b(12)+c}=182=\color{blue}{144a+12b+c.}\]
\[\color{red}{a(17)^2+b(17)+c=?}\]\[6 \le c \le 10.\]

Answer 22

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?

Answer 23

f(1) is also interesting, it gives us:\[3\le a+b+c\le4\]

Answer 24

what do you mean by /strange/?

Answer 25

is there a particular approach that is used?

Answer 26

Well, the way they did it is a lot simpler and way different from this way...they completed the square...

Answer 27

hmmm... let me think over that approach...

Answer 28

Becoz we all are doing this type of question first time.
& they had done it already.
I really appreciate @asnaseer 's work:D

Answer 29

all three equations have a turning point (minimum) at x=2. so you could use that.

Answer 30

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:\[a+b+c=4\tag{3b}\]

Answer 31

Does it become a geometric question when they complete the square?