Question

last calculas question

Answer 1

@hartnn

Answer 2

@cookiibabii93 @cookiibabii93 @zepdrix @sourwing @SithsAndGiggles @ganeshie8 @Hero @jenniferjuice @tanya123 @Luigi0210

Answer 3

@iambatman

Answer 4

@wwhitlock

Answer 5

evaluate 1/g'(1)

Answer 6

could u help me

Answer 7

Have you previously seen a formula like this?
\( (f^{-1})' (x) = \dfrac{1}{f'(f^{-1}(x))} \)

Answer 8

In our case we are looking for the derivative of the inverse of g(x).
f(x) = g(x) = x^3 + 2x - 1
f^(-1) (x) = g^(-1) (x)
Note that we know one point on the graph of the inverse function, (1, 2). x=1, g^(-1) (x) = 2. In that way, we end up with the situation that
(g^(-1))' (x) = 1/ g'( g^(-1) (2) ) = 1/ g'(1)
To complete this, we only need to compute g'(x) and plug in x=1.