SOLUTION: Hi, Please help.. a: On the one set of axes sketch the graph of {{y = f (x)}} and {{{ f(x)= f^-1(x)}}}, where domain is from 0 to infinity {{{f(x)= x^2 }}} b: find the coor

Algebra ->  Rational-functions -> SOLUTION: Hi, Please help.. a: On the one set of axes sketch the graph of {{y = f (x)}} and {{{ f(x)= f^-1(x)}}}, where domain is from 0 to infinity {{{f(x)= x^2 }}} b: find the coor      Log On


   

Question 762384: Hi, Please help..
a: On the one set of axes sketch the graph of {{y = f (x)}} and
+f%28x%29=+f%5E-1%28x%29, where domain is from 0 to infinity f%28x%29=+x%5E2+
b: find the coordinates of the points for which +f%28x%29+=+f%5E-1+%28x%29
I solve part a but I have difficulty solving part b: here is my working:
a) f+%28x%29+=+x%5E2
+y+=+x%5E2
+x+=+y%5E2
+sqrt+x+=+y I sketched the graph ok
b) +x%5E2+=+sqrt+x+
+x+=+x+ How do I find the coordinate of the point x??
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The graph of f%5E%28-1%29%28x%29 is the same as the graph of f%28x%29, but exchanging the x- and y-axes. You could also say that one graph is the reflection of the other across the y=x diagonal line.
graph%28300%2C300%2C-1%2C9%2C-1%2C9%2Cx%5E2%2Csqrt%28x%29%2Cx%29
Of course, the graph of f%5E%28-1%29%28x%29=sqrt%28x%29 is only for x%3E=0.

From x%5E2=sqrt%28x%29, squaring both sides we get
x%5E4=x for x%3E=0
(You do not get x=x in any way).
x%5E4=x --> x%5E4-x=0 --> %28x%5E3-1%29x=0
whose solutions are x=0 and x=1,
corresponding to the origin and point (1,1).