Q-f-x-e-x-x-4-is-given-put-h-x-ln-x-f-1-x-find-D-h-domain-of-h-

Question Number 173298 by mnjuly1970 last updated on 09/Jul/22

Q :

f (x) = e^{x} + x - 4 i s g i v e n

p u t : h (x) = l n (x - f^{- 1} (x))

f i n d : D_{h} = (d o m a i n o f h)

Answered by floor(10²Eta[1]) last updated on 09/Jul/22

we want : x - f^{- 1} (x) > 0 \Rightarrow x > f^{- 1} (x)

note that f^{'} (x) = e^{x} + 1 > 0 \forall x \in R

\Rightarrow f is increasing i . e ., \forall a, b \in D_{f} ∴ a > b \Rightarrow f (a) > f (b)

now back to x > f^{- 1} (x)

since f is incresing ∴ f (x) > f (f^{- 1} (x))

\Rightarrow f (x) > x \Rightarrow e^{x} + x - 4 > x \Rightarrow e^{x} > 4 \Rightarrow x > \ln 4

\Rightarrow D_{h} = {x \in R ∣ x > \ln 4}