<p>If f(x) = x^3 - x - 6 for all real numbers x, and if g is the inverse function of f, then f'(g(0))g'(0) =</p>
<p>A 0
B 1
C -1
D -6
E cubed root of 6</p>
<p>If f(x) = x^3 - x - 6 for all real numbers x, and if g is the inverse function of f, then f'(g(0))g'(0) =</p>
<p>A 0
B 1
C -1
D -6
E cubed root of 6</p>
<p>Note that f’(g(0))g’(0)=(f(g))’(0). f(g(x))=x, whose derivative is 1 wrt x, so B.</p>
<p>Wait, isn’t it asking you to evaluate it at x=0? [f(g(0))] = 0?</p>
<p>Ah, sorry. ( d(f(g(x))/dx) = 1, so regardless of where it is evaluated at, the derivative is 1. I apologize.</p>
<p>B, since it simplifies to f(g(x))=x</p>