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Mathematics 19 Online
OpenStudy (anonymous):

f(x)=\[2x^3+x-4\] Find the equation of the tangent line to f -1 (x) at the point where x = -22.

OpenStudy (anonymous):

f(22) = f'(22)(22-a) That should help you out a little bit

OpenStudy (anonymous):

Plug in x=-22. Find the y value, which is the x value for f^(-1)(x) Then plug in this "y" value into f'(x). There is the slope. Then you have the points ("y",22). Point slope form to find equation of line.

OpenStudy (anonymous):

malevolence, what's wrong with my way?

OpenStudy (anonymous):

Oh, nothing, I was just saying.

OpenStudy (anonymous):

Thank you.

OpenStudy (anonymous):

No problemmm :P

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