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Question Number 42730 by Rio Michael last updated on 01/Sep/18

G i v e n t h a t f (x) = \sqrt{1 - x} F i n d

a) D_{f} f o r t h e a r r a n g e d f o r m o f f (x)

b) f g i f f h = g (x) a n d h (x) = 3 x^{2} - 4

c) A (x) = {\begin{cases} \sqrt{1 - x}, x \neq 1 \\ x^{2}, x \neq 0 \end{cases}

f i n d A^{- 1} .

Commented by Joel578 last updated on 03/Sep/18

For (c)

if x = - 1, what function A (x) should be defined ?

A (x) = \sqrt{1 - x} or A (x) = x^{2} ?

Answered by Joel578 last updated on 02/Sep/18

(a)

D_{f} = {x \in R : x ⩽ 1}

(b)

(f . h) (x) = (3 x^{2} - 4) \sqrt{1 - x} = g (x)

(f . g) (x) = {(\sqrt{1 - x})}^{2} (3 x^{2} - 4)

= (1 - x) (3 x^{2} - 4)

= - 3 x^{3} + 3 x^{2} + 4 x - 4

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