can someone help me with this math question REALLY TOUGH.

Nov 2016
12
0
Texas
f(x) and g(x) are defined over the real number set $\small\mathbb{R}$ as follows:
g(x) = 1 - x + x² and f(x) = ax + b.
If g$\small\circ$f(x) = 16x² - 12x + 3, determine all the possible values of a and b.
 
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Jun 2014
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Earth
f(x) and g(x) are defined over the real number set $\small\mathbb{R}$ as follows:
g(x) = 1 - x + x² and f(x) = ax + b.
If g$\small\circ$f(x) = 16x² - 12x + 3, determine all the possible values of a and b.
Don't write "tothepowerof2." Use "^" for squared, for example, "x^2."

For g(x), there is a typo, as there is a gap. between 1 and x.

Is that part supposed to be "1x," "1 - x," or "1 + x," or something else? **

And, let's write (g â—¦ f)(x) instead of g â—¦ f(x).

(g â—¦ f)(x) = 16x^2 - 12x + 3


** Please clarify here what g(x) is.
 
Last edited by a moderator:
Nov 2016
12
0
Texas
Don't write "tothepowerof2." Use "^" for squared, for example, "x^2."

For g(x), there is a typo, as there is a gap. between 1 and x.

Is that part supposed to be "1x," "1 - x," or "1 + x," or something else? **

And, let's write (g â—¦ f)(x) instead of g â—¦ f(x).

(g â—¦ f)(x) = 16x^2 - 12x + 3


** Please clarify here what g(x) is.
Dear Sir,

thank you for the reply. Sorry about that could not find the key, also g(x) is 1-x.
 

topsquark

Math Team
May 2013
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1,052
The Astral plane
f(x) and g(x) are defined over the real number set $\small\mathbb{R}$ as follows:
g(x) = 1 - x + x² and f(x) = ax + b.
If g$\small\circ$f(x) = 16x² - 12x + 3, determine all the possible values of a and b.
The standard way of writing a polynomial is to list the highest power first, so \(\displaystyle g(x) = x^2 - x + 1\) and \(\displaystyle f(x) = ax + b\) and \(\displaystyle (g \circ f) (x) = 16x^2 - 12x + 3\).

\(\displaystyle (g \circ f)(x) = g(ax + b) = (ax + b)^2 - (ax + b) + 1\)

Can you finish from here?

-Dan
 
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romsek

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Sep 2015
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$g(x) = 1+x+x^2$

$f(x) = a x + b$

$(g \circ f)(x)= 16x^2 - 12x + 3$

determine all possible $a, ~b$

$(g \circ f)(x)=1+(ax+b) + (ax+b)^2 =$

$a^2 x^2+x (2 a b+a)+(b^2+b+1)$

so we have the equations

$a^2 = 16$

$(2 a b + a) = -12$

$(b^2 + b + 1) = 3$

and I leave these to you to solve.
 
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skipjack

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Dec 2006
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That needs a little adjustment to cater for the intended definition of g(x). A Google search easily finds the same method. I've tidied up the question, including supplying the missing "-".
 
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romsek

Math Team
Sep 2015
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USA
$g(x) = 1+x+x^2$

$f(x) = a x + b$

$(g \circ f)(x)= 16x^2 - 12x + 3$

determine all possible $a, ~b$

$(g \circ f)(x)=1+(ax+b) + (ax+b)^2 =$

$a^2 x^2+x (2 a b+a)+(b^2+b+1)$

so we have the equations

$a^2 = 16$

$(2 a b + a) = -12$

$(b^2 + b + 1) = 3$

and I leave these to you to solve.
oh it's 1-x+x^2 is it... of course it is

in that case

$a^2 = 16$

$(2 a b - a) = -12$

$(b^2 - b + 1) = 3$
 
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