SOLUTION: A1. Let f be a one-to-one function whose inverse function is given by the formula {{{f^(-1)(x)=x^5+2x^3+3x+1}}} (a) Compute {{{f^-1(1)}}} and {{{f(1)}}} (b) Compute the

Algebra ->  Testmodule -> SOLUTION: A1. Let f be a one-to-one function whose inverse function is given by the formula {{{f^(-1)(x)=x^5+2x^3+3x+1}}} (a) Compute {{{f^-1(1)}}} and {{{f(1)}}} (b) Compute the       Log On


   

Question 1035668: A1. Let f be a one-to-one function whose inverse function is given by the formula
f%5E%28-1%29%28x%29=x%5E5%2B2x%5E3%2B3x%2B1


(a) Compute f%5E-1%281%29 and f%281%29
(b) Compute the value of x%5B0%5D such that f%28x%5B0%5D%29+=+1.
(c) Compute the value of y%5B0%5D such that f%5E-1%28y%5B0%5D%29=1.
Thankyou for looking over! Steps and instructions will be appreciate.
Found 2 solutions by robertb, Edwin McCravy:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
(a) f%5E-1%281%29+=+1%5E5%2B2%2A1%5E3%2B3%2A1%2B1+=+7.
To find f(1), substitute f(1) into f%5E%28-1%29%28x%29
==> 1+=+f%5E-1%28f%281%29%29=%28f%281%29%29%5E5%2B2%2A%28f%281%29%29%5E3%2B3%2Af%281%29%2B1
==> ==> highlight%28f%281%29+=+0%29, since %28f%281%29%29%5E4%2B2%2A%28f%281%29%29%5E2%2B3+%3E+0.
(b) f%28x%5B0%5D%29+=+1 ==> x%5B0%5D+=+f%5E%28-1%29%281%29+=+7.
(c) f%5E-1%28y%5B0%5D%29=1 ==> f%5E-1%28y%5B0%5D%29=%28y%5B0%5D%29%5E5%2B2%2A%28y%5B0%5D%29%5E3%2B3%2Ay%5B0%5Dx%2B1+=+1
==> %28y%5B0%5D%29%5E5%2B2%2A%28y%5B0%5D%29%5E3%2B3%2Ay%5B0%5D+=+0 ==> %28y%5B0%5D%29%28+%28y%5B0%5D%29%5E4%2B2%2A%28y%5B0%5D%29%5E2%2B3+++++++++++++++%29+=+0 ==> highlight%28y%5B0%5D+=+0%29. (This is basically the same computation we had in the latter half of part (a).

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
First I'll teach you a little bit about inverse functions.
Stuff you should know before you try to do inverse problems:





This is the graph of the inverse of f(x), which we call f-1(x). 

The inverse of a function is the graph made by interchanging
the x-coordinates with the y-coordinates of the points on
the graph.

That is to say that if (a,b) is a point on f(x), then
(b,a) is a point on f-1(x).

If the equation for f-1(x) were simpler we could
calculate it and graph it.  But we can't do that here because
we don't know how to solve a 5th degree polynomial for x.

I will find a few points on f(x), by finding
a few points on f-1(x) and reversing the coordinates.

   x   f-1(x)    point on f-1(x)   point on f(x)
------------------------------------------------
  -1   -5.00       (-1,-5.00)       (-5.00,1)
-0.9   -3.75       (-0.9,-3.75)     (-3.75,-0.9)     
-0.8   -2.75       (-0.8,-2.75)     (-2.75,-0.8)
-0.7   -1.95       (-0.7,-1.95)     (-1.95,-0.7)
-0.5   -0.78       (-0.5,-0.78)     (-0.78,-0.5)
 0.0    1.00       (0,1.00)         (1.00,0)
 0.3    1.96       (0.3,1.96)       (1.96,0.3)
 0.5    2.78       (0.5,2.78)       (2.78,0.5)
 0.7    3.95       (0.7,3.95)       (3.95,0.7)
 0.8    4.75       (0.8,4.75)       (4.75,0.8)

Here are those 10 points of f(x) plotted:



Now we'll look at your questions:



(a) Compute f-1(1) and f(1)

To find f-1(1), substitute x=1 in




So f-1(1) = 7

That means the point (1,7) is on the graph of f-1(x). 

We want to find f(1).

We notice that since  ends with a 1.
Then if x=0 then 

So the point (0,1) is on the graph of f-1(x),
so the point (1,0) is on the graph of f(x), so

f(1) = 0 


(b) Compute the value of x%5B0%5D such that f%28x%5B0%5D%29+=+1.

That means (x0,1) must be on f(x), so its reverse,
(1,x0) must be on f-1(x), and since f-1(1) = 7, 
(1,7) is on f-1(x) and that means (7,1) is on f(x), so
therefore f(7)=1 so x0 = 7

 (c) Compute the value of y%5B0%5D such that f%5E-1%28y%5B0%5D%29=1.

That means (y0,1) must be on f-1(x), so its reverse,
(1,y0) must be on f(x), and since f(1) = 0, 
(1,0) is on f(x) and that means (0,1) is on f-1(x), so
therefore f-1(0)=1 so y0 = 0.

Edwin