SOLUTION: If {{{ g(x) = 3 + x + e^x }}} , find {{{ g^-1(4) }}} Me and my friend are so stuck on this problem. We have no clue how to solve it. We switched the x values and y values but go

Algebra ->  Test -> SOLUTION: If {{{ g(x) = 3 + x + e^x }}} , find {{{ g^-1(4) }}} Me and my friend are so stuck on this problem. We have no clue how to solve it. We switched the x values and y values but go      Log On


   

Question 217256: If +g%28x%29+=+3+%2B+x+%2B+e%5Ex+ , find +g%5E-1%284%29+
Me and my friend are so stuck on this problem. We have no clue how to solve it. We switched the x values and y values but got stuck on:
+x-3+=+y+%2B+e%5Ey+
Could you please help us out! Thanks in advance. =)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
You've done everything correctly so far. Now we're looking for g%5E-1%284%29. If we replace x by 4 in your equation for the inverse we get:
4-3=y%2Be%5Ey
which simplifies to
1=y%2Be%5Ey
which leaves us with figuring out y. Other than some logic, combined with trial and error, the only way to find y that I can suggest is to look at the graph of g(x). (I'd suggest looking at the graph of the inverse but Algebra.com's graphing software will not work on x-3=y%2Be%5Ey.)
graph%28600%2C+600%2C+-8%2C+8%2C+-8%2C+8%2C+3%2Bx%2Be%5Ex%29
Since the function's x's are the inverse's y's and vice versa and since we are looking for g%5E%28-1%29%284%29, we can try to find where g(x) = 4. And from the graph it appears that g(0) = 4. And we can verify this by substituting 0 in for x and finding g(0) (which does indeed turn out to be 4). Since g(0) = 4 then g%5E%28-1%29%284%29+=+0.