Suppose f is a one-to-one, differentiable function and its inverse function f^−1 is also differentiable. One can show, using implicit differentiation (do it!), that

(f^−1)′(x)=1 / f′(f^−1(x))

Find (f^−1)′(−6) if f(−1)=−6 and f′(−1)=3/5.

(f^−1)′(−6)=

you have g(x) = f^-1(x)

Since f(-1) = -6, g(-6) = -1
(f^-1(-6))' = 1/g'(-1) = 1/(3/5) = 5/3

Well, isn't this a fun little problem! Let's dive into it, shall we?

Now, we know that (f^−1)′(x) = 1 / f′(f^−1(x)). So all we need to do is plug in the given values and see what happens!

First, we have f(−1) = −6. This means that f^−1(−6) = −1. Pretty straightforward, right?

Next, we have f′(−1) = 3/5. Now, we need to find f′(f^−1(−6)). But since f^−1(−6) = −1, we can replace f^−1(x) with −1 in our expression. So we get f′(f^−1(−6)) = f′(−1).

Now, all we need to do is substitute these values into the formula (f^−1)′(x) = 1 / f′(f^−1(x)). We get:

(f^−1)′(−6) = 1 / f′(−1) = 1 / (3/5) = 5/3.

So there you have it! (f^−1)′(−6) = 5/3. And now that we're done, can I interest you in some more math-related jokes?

To find (f^−1)′(−6), we will use the formula:

(f^−1)′(x)=1 / f′(f^−1(x))

Given that f(−1)=−6 and f′(−1)=3/5, we can substitute these values into the formula.

To find f^−1(x), we can switch the roles of x and y in the equation f(x) = y and solve for x, which gives us the inverse function.

Given that f(−1)=−6, we can rewrite this as:

f(x) = y
f(−1) = −6

Now, let's find f^−1(x):

y = f(x)
−6 = f(−1)

Switching the roles of x and y, we have:

x = f^−1(y)
−1 = f^−1(−6)

So, f^−1(−6) = −1.

Now, let's find (f^−1)′(−6):

(f^−1)′(−6) = 1 / f′(f^−1(−6))

Substituting f^−1(−6) = −1 and f′(−1)=3/5 into the formula, we get:

(f^−1)′(−6) = 1 / (3/5)

To divide by a fraction, we can multiply by its reciprocal:

(f^−1)′(−6) = 1 * (5/3)

Multiply the numbers:

(f^−1)′(−6) = 5/3

Therefore, (f^−1)′(−6) = 5/3.

To find (f^−1)′(−6), we first need to find f^−1(x) by solving for x in terms of f(x). Given that f(−1)=−6, we know that f^−1(−6)=−1.

Now, let's differentiate both sides of the equation f(x)=−6 with respect to x:

d/dx [f(x)] = d/dx [−6]

Using the chain rule, we have:

f′(f^−1(x)) * (f^−1)′(x) = 0

Since f′(−1)=3/5, we substitute f^−1(x)=−1 and f′(f^−1(x)) = f′(−1) into the equation:

(3/5) * (f^−1)′(x) = 0

Now we can solve for (f^−1)′(x):

(f^−1)′(x) = 0 / (3/5)
= 0 * (5/3)
= 0

Therefore, (f^−1)′(−6) = 0.