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:
I don't understand percentages
Answer:
x = -8
Step-by-step explanation:
1/4x+2=-5/8x-5
Add 5/8 x to each side
1/4x + 5/8 x+2=-5/8x + 5/8x-5
1/4x + 5/8x +2 = -5
Subtract 2 from each side
1/4x + 5/8x +2-2 = -5-2
1/4x + 5/8x = -7
Get a common denominator of 8
1/4 *2/2 x + 5/8x = -7
2/8x + 5/8x = -7
7/8x = -7
Multiply each side by 8/7
8/7 * 7/8x = -7*8/7
x = -8
