Given #f(x) = (3-2x) / (2x+1)# and #f(g(x)) = 7 - 3x# how do you find g(x)?

Question

Elias Ackerman · Accepted Answer

undefined To find $ g(x) $, we need to express $ f(g(x)) $ in terms of $ g(x) $ and then compare it with the given expression $ f(g(x)) = 7 - 3x $.

Given $ f(x) = \frac{{3 - 2x}}{{2x + 1}} $ and $ f(g(x)) = 7 - 3x $, we can express $ f(g(x)) $ as:

$$ f(g(x)) = \frac{{3 - 2g(x)}}{{2g(x) + 1}} = 7 - 3x $$

By comparing the numerators and denominators, we can write:

$$ 3 - 2g(x) = 7 - 3x $$
$$ 2g(x) + 1 = -3x $$

Now, solve this system of equations for $ g(x) $:

From the second equation, we have:

$$ 2g(x) = -3x - 1 $$
$$ g(x) = \frac{{-3x - 1}}{2} $$

So, $ g(x) = -\frac{{3x}}{2} - \frac{1}{2} $.

Theodore Amidon · Answer

#g(x) = (4 - 3x) / (-16 + 6x)#

#f(g(x))# can be computed by plugging #g(x)# for every occurence of #x# in #f(x)#.

Even though we don't know #g(x)# yet, we can still do this:

#f(x) = (3 - 2x) / (2x + 1) " "=> " " f(g(x)) = (3 - 2 g(x)) / (2 g(x) + 1)#

We also know that #f(g(x)) = 7 - 3x#, so we have:

#(3 - 2 g(x)) / (2 g(x) + 1) = 7 - 3x#

Let me write #g# instead of #g(x)# for better readability:

#(3 - 2g)/(2g + 1) = 7 - 3x#

Now, you need to solve this equation for #g#:

... multiply both sides with #(2g+1)#...

#<=> 3 - 2g = (7 - 3x) * (2g + 1)#

#<=> 3 - 2g = (7 - 3x) * 2g + (7 - 3x)#

Bring all products that include #g# to the left side and everything else to the right side. So, subtract #(7 - 3x) * 2g# on both sides, and subtract #3# on both sides:

#<=> - 2g - (7 - 3x) * 2g = (7 - 3x) - 3#

... factorize #g# on the left side...

#<=> (-2 - 14 + 6x) * g = 4 - 3x#

#<=> (-16 + 6x) * g = 4 - 3x#

... divide both sides by #(-16 + 6x)#...

#<=> g = (4 - 3x)/(-16 + 6x) = (4 - 3x)/(2(-8 + 3x))#

Thus, we have

#g(x) = (4 - 3x) / (-16 + 6x)#

It might be a good idea to test if the calculation was correct. To do so, compute #f(g(x))#:

#f(g(x)) = f((4 - 3x) / (-16 + 6x)) #

#= (3 - 2 * (4 - 3x) / (2(-8+ 3x)))/(2 * (4 - 3x) / (2(-8 + 3x)) + 1) #

#= (3 - (4 - 3x) / (-8 + 3x))/( (4 - 3x) /(-8+ 3x) + 1) #

#= ((3(-8 + 3x) - (4 - 3x))/(-8 + 3x)) / ((4 - 3x + (-8 + 3x))/(-8 + 3x))#

#= (3(-8 + 3x) - (4 - 3x)) / (4 - 3x + (-8 + 3x)) = (-28 +12x) / (-4)#

#= 7 - 3x#