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.