Since f(x) is (strictly) increasing, we know that it is one-to-one and has an inverse f^(-1)(x). Then we can apply the inverse function theorem. Suppose f(a) = b and a = f^(-1)(b). By definition of inverse function, we have
f^(-1)(f(x)) = x
Differentiating with the chain rule gives
(f^(-1))'(f(x)) f'(x) = 1
so that
(f^(-1))'(f(x)) = 1/f'(x)
Let x = a; then
(f^(-1))'(f(a)) = 1/f'(a)
(f^(-1))'(b) = 1/f'(a)
In particular, we take a = 2 and b = 7; then
(f^(-1))'(7) = 1/f'(2) = 1/5
Answer:A 12/13
Step-by-step explanation:
Answer:
2 hours
Step-by-step explanation:
monday = 6 hrs
tuesday = 4 hrs
6 - 4 = 2
2 hours
Im pretty sure it would be 0.2