<p>f(g(x)) = 4x^2 - 8x and f(x) = x^2 - 4</p>
<p>then g(x) = ???</p>
<p>How do you do this algebriacly?</p>
<p>Thanks</p>
<p>f(g(x)) just means put g(x) into the input of f(x) </p>
<p>So you get the equation g(x)^2 - 4 = 4x^2 - 8x </p>
<p>Solve for g(x)</p>
<p>I solved for g(x) but I got the square root of a quadratic equation. </p>
<p>Is there no faster way?</p>
<p>g(x)^2 = 4x^2 - 8x + 4 = 4(x^2 - 2x + 1) = (2(x-1))^2</p>
<p>Note that we must be particularly careful when taking the “square root” of both sides. The statement a^2 = b^2 does not imply a = b, it only implies a = ±b. In this case we get</p>
<p>g(x) = ± 2(x - 1)</p>
<p>Even then, the above is not quite correct. g can consist of arbitrary combinations of the above (i.e. g(x) = 2x-2 on some intervals and -2x+2 on the rest). However for the purposes of the SAT, this is likely not the case, so g(x) = 2x - 2 or g(x) = -2x + 2 should work.</p>
<p>First of all, this question would not appear on the real SAT.</p>
<p>Now, as MITer94 said, there is an infinite number of possible answers if we consider piecewise functions.
There are only six continuous functions g(x) with four distinct graphs:
g(x) = 2x - 2,
g(x) = -2x + 2
g(x) = |2x - 2|, g(x) = |-2x + 2|
g(x) = -|2x - 2|, g(x) = -|-2x + 2|</p>
<p>Graphing interpretation on TI-8x for the first function.</p>
<p>Y1 = g(x)
Y2 = f(x)
Y3 = f(g(x))
that is</p>
<p>Y1 =2x-2
Y2 =x²-4
Y3 =Y2(Y1)</p>
<p>Turn off the entries Y1 and Y2, so that their graphs do not appear when the composite function Y3 is graphed.
Set the window Xmin=-1, Xmax=3, Ymin=-5, Ymax=6.
Graph Y3.
Now graph Y4 = 4x² - 8x setting either Thick Line, or Path, or Animated style.
Graphs of Y3 and Y4 are identical,
i.e. f(g(x)) = 4x² - 8x for f(x)=x²-4 and g(x)=2x-2.</p>