The probability would be 2/6 or 1/3
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:
b). (0, 4)
if you need an explanation, comment, and I will put one in for you :)
Answer:
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Step-by-step explanation:
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