Question

Let h be a continuous, differentiable one to one function such that g(3) = 7, g (7) = 3, g
(3) = 2, and g
(7) = 4.
a) Find (g^-1)(3) and (g^-1)'(3)

Thanks,
Let h be a continuous, differentiable one to one function such that g(3) = 7, g (7) = 3, g
(3) = 2, and g
(7) = 4.
a) Find (g^-1)(3) and (g^-1)'(3)

Thanks,
@Mathematics

Answer 1

\[g(x)=y\]\[g^{-1}(y)=x\]
but if g(3)=7 and g(3)=2 as your problem states this is not a one-to-one function. I sense a typo...
You also start talking about some function "h" which does not appear in the problem. Please clarify.

Answer 2

Ahhh I'm sorry, its g'(-7) =4 and g'(3)=2

Answer 3

So instead it should be such that g(3) = -7, g (-7) = 3, g'(3) = 2, and g'(-7) = 4.

Answer 4

\[g(x)=y\]\[g^{−1}(y)=x\]here we have
g(3)=-7
so x=3, y=-7
so it looks like\[ g^{-1}(-7)=3\](seems strange to me since we were given that g(-7)=3, but whatever...)

for g'(3)=2
x=3, y=2
so \[(g^{-1})'(2)=3\]
for g'(-7)=4\[(g^{-1})'(4)=-7\]
I see no way to get (g^-1)'(3) from the given information. I may have forgotten some trick, but I sense perhaps another typo in the question?