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))
If f(4)=5 and f′(4)=2/3, find (f−1)(5).
(f−1)′(5)=
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f^-1(x) for f inverse
To find (f⁻¹)'(5), we can use the formula (f⁻¹)'(x) = 1 / f'(f⁻¹(x)).
Given that f(4) = 5 and f'(4) = 2/3, we need to find f⁻¹(5).
To solve this, we'll use the fact that f and f⁻¹ are inverses of each other. So if we apply f⁻¹ to the equation f(4) = 5, we should get the equation f⁻¹(5) = 4.
Now let's use the formula (f⁻¹)'(x) = 1 / f'(f⁻¹(x)).
Plugging in x = 5, we have (f⁻¹)'(5) = 1 / f'(f⁻¹(5)).
Since f⁻¹(5) = 4, we have (f⁻¹)'(5) = 1 / f'(4).
Given that f'(4) = 2/3, we have (f⁻¹)'(5) = 1 / (2/3).
To divide by a fraction, we can multiply by its reciprocal. So (f⁻¹)'(5) = 1 * (3/2).
Multiply straight across, and we get (f⁻¹)'(5) = 3/2.
Therefore, (f⁻¹)'(5) equals 3/2.