Answer:
(a) 1/9
Step-by-step explanation:
I find these easier to understand by remembering that the graph of an inverse function is the graph of the original function reflected across the line y=x.
The point (2, f(2)) is the point (2, 7). The slope of the function at that point is ...
f'(2) = 3x² -3 = 3(2² -1) = 9
When the graph is reflected across the line y=x, the slope at the corresponding point on the inverse function, (7, 2), will be the reciprocal of that slope, 1/9.
[tex](f^{-1})'(f(2))=\dfrac{1}{f'(2)}=\dfrac{1}{9}[/tex]
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In the attached graph, the red curve is the original cubic, f(x). The blue curve is the inverse relation. (You will notice it is not a function.) The corresponding points (2, f(2)) = (2, 7) and (f(2), 2) = (7, f⁻¹(7)) are shown. The tangent line is drawn with a slope of 1/9, so you can see that is the appropriate choice for (f⁻¹)'(7).
