Let f (x) = x ^ 2 + X (x)

Let f (x) = x ^ 2 + X (x)

The solution of F (x) = x ^ 2 + x = 2 is x = - 2, and the inverse function f (2) = - 2

Let y = f (x) be the inverse function of x = φ (y), and f (10) = 2, f '(10) = 5, find φ' (2)

Let y = f (x) be the inverse function of x = φ (y), and f (10) = 2, f '(10) = 5, find φ' (2)
∵ y = f (x) is the inverse function of x = φ (y), and f (10) = 2, ∵ φ (2) = 10, φ′ (2) = 1 / F ′ (10) = 1 / 5

If FX = 10 ^ x, the inverse function of FX is F-1 X. if a is greater than 1 and B is greater than 1, then F-1 (a) / F-1 (b)=
It's urgent!

Because FX = 10 ^ X
So when x is greater than 0, F-1 x = lgx
And because a is greater than 1, B is greater than 1
therefore
f-1(a)/f-1(b)
=lga/lgb

Let f ^ - 1 (x) be the inverse function of the function f (x) = 0.5 (a ^ x-a ^ - x) (a is greater than 1), then if f ^ - 1 (x) is greater than 1, the value range of X is?

Because the range of f ^ - 1 (x) is the domain of F (x), we can find the range of X by f ^ - 1 (x) > 1, that is to know x > 1 and find the range of F (x), because f (x) = 0.5 (a ^ x-a ^ - x) f (x) '= 0.5 (a ^ x ㏑ a + A ^ - x ㏑ a). When 0 < a < 1, f' (x) < 0, f (x) is a decreasing function, then x > 1, f (x) < f (1) = 0.5 (...)