<p>If f(x)=2x((Square root)x^2+1), f(g(x))=x, then g(1)=...
a .207
b .455
c .707
d 1.207
e 2.828</p>
<p>scroll down for answer</p>
<p>B. </p>
<p>How the heck do you get the answer? thanks</p>
<p>square root around the x^2 + 1, right?</p>
<p>g(1) = (something) <=> f(g(1)) = f(something) = 1</p>
<p>to solve f(something) = 1, graph f(x) and find where it equals 1</p>
<p>It isn't that bad if you have a TI-89.<br>
Since f(g(x)) = x, that means g(x) is the inverse function of f(x). So you then switch the x and y to get that. </p>
<p>x = 2y * (sqrt(y^2 +1))</p>
<p>Then I used the solver on my TI-89 to solve for y.
I plugged in 1 for x.</p>
<p>I got 0.44509 as my answer.</p>
<p>To extend on Pentasa's post, if you want to solve this without graphing, you need to solve</p>
<p>2y * sqrt( y^2 + 1) = 1 where y=g(1)
Square both sides, you get
4y^2 (y^2 + 1) = 1
For simplicity, set z=y^2; you now have a quadratic equation in z.</p>
<p>Solve 4z^2 + 4z - 1 = 0 to get z= -1.202 or + 0.207
Since z = y^2, it must be >=0, so use z = 0.207 = g(1)^2
from which g(1) = sqrt(0.207) = 0.455</p>
<p>i understand how to solve it by algebra, but I don't get how you set up the problem. I know that to find the inverse of a function, you switch the xs and the ys, then you solve for x. then you plug in the "1" into the inverse function...is that right? thanks for all the help...i appreciate it.</p>
<p>okay ive got it:
basically set 2x((Square root)x^2+1) = x
seems too simple, but you get the right answer...</p>
<p>i just plugged in the answer choices...lol i got the right answer and it took me less than 30 seconds.</p>