<span>7(-3 + 2u) simplified is </span>
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
Step-by-step explanation:
Given
The function is 
for 
, we will check the values of 
and 
Take 3 and 4
Similarly,
Thus, 
Also, 
is 1 i.e. a constant value. So, the function is increasing linearly for the given interval.
Answer:
1/42,042 or 2.38×10^-5
Step-by-step explanation:
Please kindly check the attached files for explanation
