SOLUTION: Please help me with this problem: Suppose {{{f(x) = x^3}}} + 2x and {{{f^-1}}} is the inverse of {{{f}}}. Evaluate {{{f^-1(3)}}} and {{{(f^-1)}}}'{{{(3)}}}.

Algebra ->  Functions -> SOLUTION: Please help me with this problem: Suppose {{{f(x) = x^3}}} + 2x and {{{f^-1}}} is the inverse of {{{f}}}. Evaluate {{{f^-1(3)}}} and {{{(f^-1)}}}'{{{(3)}}}.      Log On


   

Question 1034949: Please help me with this problem:
Suppose f%28x%29+=+x%5E3 + 2x and f%5E-1 is the inverse of f. Evaluate f%5E-1%283%29 and %28f%5E-1%29'%283%29.
Found 2 solutions by solver91311, robertb:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


I assume you are asking for both the inverse function and the first derivative of the inverse function.

Let

then

then

which is to say:



Swap the variables:



then



The latter being more convenient for using the power rule to find the first derivative.



Plug in the value 3 for x in each case.

John

My calculator said it, I believe it, that settles it

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+x%5E3+%2B2x+ ==> f%28f%5E-1%28x%29%29+=+x+=+%28f%5E-1%28x%29%29%5E3+%2B+2f%5E-1%28x%29+ after plugging in the inverse f%5E-1%28x%29 into the original equation.
Now differentiate implicitly wrt x.
==>
Now f%28x%29+=+x%5E3+%2B2x+ ==> df%28x%29%2Fdx+=+3x%5E2+%2B2+%3E+0+ for all real x hence f(x) increasing ==> f(x) is one-to- DISABLED_event_one= => f(x) has an inverse function. Incidentally, f(1) = 3, and so f%5E-1%283%29+=+1.

==> 1+=++%283%2A%28f%5E-1%283%29%29%5E2+%2B+2%29%28df%5E-1%283%29%2Fdx%29++
==> 1+=+%283%2A1%5E2+%2B+2%29%28df%5E-1%283%29%2Fdx%29+=+5%28df%5E-1%283%29%2Fdx%29
Therefore, df%5E-1%283%29%2Fdx+=+1%2F5