SOLUTION: Please help me with this: The function f(x) = {{{x^3}}} + x is a one-to-one function, and thus its inverse {{{f^-1}}} is also a function. Find the equation of the tangent line whi

Algebra ->  Test -> SOLUTION: Please help me with this: The function f(x) = {{{x^3}}} + x is a one-to-one function, and thus its inverse {{{f^-1}}} is also a function. Find the equation of the tangent line whi      Log On


   

Question 1033081: Please help me with this:
The function f(x) = x%5E3 + x is a one-to-one function, and thus its inverse f%5E-1 is also a function. Find the equation of the tangent line which can be drawn to the graph of f%5E-1 at the point (2,1).
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
If f%28x%29+=+x%5E3+%2B+x, then plugging in f%5E-1%28x%29 into the former would lead to
f%28f%5E-1%28x%29%29+=+x+=+%28f%5E-1%28x%29%29%5E3+%2B+f%5E-1%28x%29
==> %28f%5E-1%28x%29%29%5E3+%2B+f%5E-1%28x%29+=+x
==> 3%28f%5E-1%28x%29%29%5E2%2A%28df%5E-1%28x%29%2Fdx%29+%2B+df%5E-1%28x%29%2Fdx+=+1 after implicit differentiation
==> %283%28f%5E-1%28x%29%29%5E2+%2B+1%29%2A%28df%5E-1%28x%29%2Fdx%29+=+1 after factoring...
Now one particular point in the graph of f%5E-1%28x%29 is (2,1).
==> %283%28f%5E-1%282%29%29%5E2+%2B+1%29%2A%28df%5E-1%282%29%2Fdx%29+=+1
==> %283%2A%281%29%5E2+%2B1%29%2A%28df%5E-1%282%29%2Fdx%29+=+1, or 4%28df%5E-1%282%29%2Fdx%29+=+1
==> df%5E-1%282%29%2Fdx+=+1%2F4, which is also the slope of the tangent line.
==> the equation of the tangent line to f%5E-1%28x%29 at the point (2,1) is y+-+1+=+%281%2F4%29%28x+-+2%29, or y+=+x%2F4+%2B1%2F2.



As what would other tutors would point out later, the derivative at a point on the graph of f%5E-1%28x%29 is equal to the reciprocal of the derivative of the function f(x) at the inverse point, on condition that f'(x) at the inverse point is NOT zero.