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f(x)=\[2x^3+x-4\] Find the equation of the tangent line to f -1 (x) at the point where x = -22.
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f(22) = f'(22)(22-a) That should help you out a little bit
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.
malevolence, what's wrong with my way?
Oh, nothing, I was just saying.
Thank you.
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No problemmm :P
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