Question

Please HELP!! Question Below

Answer 1

AUUUUMMM!!!!

Answer 2

We are interested in f(10).
When x = 10, which domain should we choose?

Answer 3

the third equation, where x>0

Answer 4

Yes. Substitute x = 10 and isolate |f(10)| and then solve for f(10) without the absolute bars.

Answer 5

but how, i end up with \[\left| f(10) \right| + g(10) = h(10), or h(10) - g(10) = \left| f(10) \right| \]

Answer 6

correct. And whenever you get an equality with an absolute bar such as |y| = a, it means y = +a or y = -a. Apply the same rule here.

Answer 7

so \[f(10) = h(10) - g(10), or f(10) = g(10) - h(10)\]

Answer 8

correct. Now which one has the lower value: h(10) - g(10) or g(10) - h(10) ?

Answer 9

ummmm, not sure

Answer 10

i think its the first but??

Answer 11

i mean the second

Answer 12

Hint: Look at the equation you got for |f(10)| with the absolute bars.

Answer 13

so, the second is smaller

Answer 14

Yes, g(10) - h(10) is the least possible value of f(10).

Answer 15

i need a numerical value though

Answer 16

it is definitely an integer

Answer 17

and it has to be negative (apparently), unless it is 0

Answer 18

Let me look at the other two domains and see if we can look at continuity of the function and figure out what g(10) - h(10) might be.

Answer 19

ok, thanks

Answer 20

|f(x)| + g(x) = 4x for x <= -2
|f(x)| + g(x) = -2x^2 for -2 < x <= 0

We are told that f(x), g(x) and h(x) are all polynomials.
If we look at the left hand side of the above two equations they are the same.
But on the right hand side they are different! How can that be?
It means f(x) goes from a positive to a negative at x = -2 OR from a negative to a positive at x = -2 causing |f(x)| to be different to the left of x = -2 and to the right of x = -2.
With me so far?

Answer 21

f(x) + g(x) = 4x for x <= -2 AND -f(x) + g(x) = -2x^2 for -2 < x <= 0
OR
-f(x) + g(x) = 4x for x <= -2 AND f(x) + g(x) = -2x^2 for -2 < x <= 0

In either case, subtract the second equation from the first in each case:
f(x) - (-f(x))= 4x + 2x^2 ==> 2f(x) = 4x + 2x^2 ==> f(x) = 2x + x^2
OR
-f(x) - (f(x))= 4x + 2x^2 ==> -2f(x) = 4x + 2x^2 ==> f(x) = -(2x + x^2)

f(10) = 2(10) + 10^2 = 20 + 100 = 120
OR
f(10) = -(2(10) + 10^2) = -(20 + 100) = -120

The least possible value of f(x) is -120 and it happens if f(x) = -(2x + x^2)

Answer 22

Last line it should be f(10) in the place of the first f(x):
The least possible value of f(10) is -120 and it happens if f(x) = -(2x + x^2)

Answer 23

wait up though, how can you compare the equations with dif domains

Answer 24

That is what was initially throwing me off.
But they tell you f(x), g(x), h(x) are polynomials and we should have paid more attention to that statement. Domains of polynomials are (-infinity, infinity).
So the reason they have 3 different domains even though they are additions of polynomials which itself is a polynomial is because of the absolute bars. That changes the behavior of the polynomial when it changes sign. The absolute bar will flip the graph below the x-axis to above the x-axis. Thus the need for different domains arise.

Answer 25

It is a single function f(x) throughout ALL domains from -infinity to +infinity.
But because f(x) became |f(x)|, the negative portion of f(x) got flipped to the positive side which apparently happens twice for this f(x) giving rise to three domains.

Answer 26

but why is it that one needs three equations to represent this? shouldn't it just be one?

Answer 27

Look at the graph of f(x) and |f(x)| we determined for this problem:
https://www.diigo.com/item/image/4xtew/dnxu?size=o

Answer 28

so why 3 equations? theres only 2 parabolas

Answer 29

If we just added f(x) + g(x), it will be just one polynomial with the domain -inf < x < +inf. But here |f(x)| is added to g(x). So when f(x) < 0, |f(x)| = -f(x)
when f(x) > 0, |f(x)| = f(x).

Here since f(x) changes sign twice (at x = -2 and at x = 0), there are 3 domains.

In fact you can even say h(x) = 4x (based on the first function because f(x) goes to negative in the third domain as it was in the first domain and so the left hand side and right hand side must be identical.

Answer 30

this is really beyond anything i've learned so far.. I can't grasp the domain thing

Answer 31

i know it means the set of numbers in the solution set, but why, then are three domains necessary? does the equation f(x) change? or does the domain change because of the g(x)

Answer 32

This is indeed a tricky problem but no need to worry.

The purpose behind this problem is they want you to figure out what f(x) is based on the given data. But IF you know f(x) and g(x) there is really no need for 3 domains. You can still take the absolute value of f(x), add it to g(x) and say the domain of this new function is -infinity < x < infinity.

But here they don't give you f(x). Instead they give you the behavior of |f(x)| + g(x) in three different from which they want you to figure out f(x).

Answer 33

" three different domains from ..."

Answer 34

ohhhhhhhhhhhhh. Got it. but why can you add equations with different domains without a problem?

Answer 35

It is the same polynomial f(x) and g(x) and their sum in all 3 domains.
Essentially the new function is |f(x)| + g(x) in all three domains.
But by artificially splitting the domain into three parts they are giving you enough hints to figure out f(x).

Answer 36

They do say in the beginning, f(x), g(x) and h(x) are all polynomials.

Answer 37

f(x) and g(x) are polynomials that HAPPENS to satisfy the following equations: ....
Figure out what the polynomial f(x) might be. (they are asking this indirectly).

Answer 38

i think i got it now, Thanks! And by the way, the end result does come out to -120 correct?

Answer 39

yes. -120 is the least possible value of f(10).

Answer 40

THAAANKS!!! go AUM!!!!!! bye

Answer 41

You are welcome.