If g is an increasing function such that g(2)=8 and g'(2)=5, calculate the inverse of g' of 8

Question

If g is an increasing function such that g(2)=8 and g'(2)=5, calculate (g^-1)'8

Answer 1

g is a function with g(2) = 8.

g is increasing, so it has an inverse function g^-1.

The inverse function g^-1 has g^-1(8) = 2, exactly the opposite of our original g(2) = 8. That's what an inverse function does! ("g turns 2 into 8; g^-1 turns 8 into 2.")

We're given that g'(2) = 5. (No big surprise that this derivative g'(2) is a positive number (namely 5), since they told us that g is an increasing function.)

Now they want us to find (g^-1)'(8).

There's a formula for finding (g^-1)', which is the derivative of the inverse function (g^-1).

The formula is

(g^-1)'(y) = 1/g'(g^-1(y)) for every number y in the domain of g'

In our case, y = 8 because they're asking us to find (g^-1)'(8).

Plugging y=8 into our formula (g^-1)'(y) = 1/g'(g^-1(y)), we get

(g^-1)'(8) = 1/g'(g^-1(8))

Fortunately, we already know that g^-1(8) = 2 (see the first paragraph), so

(g^-1)'(8) = 1/g'(g^-1(8)) = 1/g'(2)

and they told us that g'(2) = 5 (no big surprise), so

(g^-1)'(8) = 1/g'(g^-1(8)) = 1/g'(2) = 1/5

So (g^-1)'(8) = 1/5 is the answer.

Answer 2

Note: (g^-1)'(8) = (g^-1)'(g(2)) (sub 8 = g(2))

We're also given g'(2) = 5, so we want to incorporate this info.

From inverse function definition, g^-1(g(x)) = x

Take derivative wrt x both sides, using chain rule: d[g^-1(g(x))]/dx = d(x)/dx

--> (g^-1)'(g(x)). g'(x) = dx / dx = 1

--> (g^-1)'(g(2)). g'(2) = 1 (sub x = 2)

--> (g^-1)'(8). (5) = 1 (given from the problem: g(2) = 8, g'(2) = 5)

--> (g^-1)'(8) = 1/5