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)′(−4) if f(5)=−4 and f′(5)=2/7.
(f^−1)′(−4)=
Why did the function go to therapy?
Because it had an identity crisis! Now, let's solve the problem.
We know that f(5) = -4 and f'(5) = 2/7. We want to find (f^−1)'(-4).
Using the formula, we have:
(f^−1)'(-4) = 1 / f'(-4)
But we don't know f'(-4), so we can't directly substitute it in. However, since f is one-to-one, the point (-4, 5) must lie on the graph of f^-1. So, we can use this information to find f'(-4).
The derivative of f^-1 at x = -4 is the reciprocal of the derivative of f at x = 5:
(f^−1)'(-4) = 1 / f'(5) = 1 / (2/7) = 7/2
Therefore, (f^−1)'(-4) = 7/2.
To find (f^−1)′(−4), we can use the formula:
(f^−1)′(x) = 1 / f′(f^−1(x))
Given that f(5) = −4 and f′(5) = 2/7, we need to find f^−1(−4) to compute (f^−1)′(−4).
To find f^−1(−4), we can use the fact that f^−1 undoes the transformation performed by f. Since f(5) = −4, it implies that f^−1(−4) = 5.
Now, substituting f^−1(−4) = 5, f′(5) = 2/7, and x = −4 into the formula, we get:
(f^−1)′(−4) = 1 / f′(f^−1(−4))
= 1 / f′(5)
= 1 / (2/7)
= 7/2
Therefore, (f^−1)′(−4) = 7/2.
To find (f^−1)′(−4), we can use the given information about f and its derivative. Let's use the formula we derived using implicit differentiation:
f(^−1)′(x) = 1 / f′(f^−1(x))
We are given that f(5) = -4 and f′(5) = 2/7. We want to find (f^−1)′(−4), so we plug -4 into the formula for x:
(f^−1)′(-4) = 1 / f′(f^−1(-4))
Now, we need to find f^−1(-4). Since f(5) = -4, it means that the input x=5 corresponds to the output y=-4. Therefore, f^−1(-4) = 5.
Now we can substitute this value into the formula:
(f^−1)′(-4) = 1 / f′(5)
Given that f′(5) = 2/7:
(f^−1)′(-4) = 1 / (2/7)
To divide by a fraction, we can multiply by its reciprocal:
(f^−1)′(-4) = 1 * (7/2)
Multiplying 1 by 7/2:
(f^−1)′(-4) = 7/2
So, (f^−1)′(−4) = 7/2.