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Question Number 93898 by mathmax by abdo last updated on 16/May/20

l e t f (x) = 2 \sqrt{3 - x} a n d g (x) = x^{2} - 2 x + 5

1) c a l c u l a t e f o g (x) a n d d e t e r m i n e D_{f o g}

2) c a l c u l a t e \int f o g (x) d x

3) c a l c u l a t e \int \frac{f^{- 1} (x)}{f (x)} d x a n d \int \frac{f^{- 1} o g (x)}{f o g (x)} d x

Commented by mathmax by abdo last updated on 17/May/20

s o r r y g (x) = x^{2} - 2 x - 5

Answered by Kunal12588 last updated on 16/May/20

1 ⟩ f o g (x) = 2 \sqrt{3 - x^{2} + 2 x - 5} = 2 \sqrt{2 x - x^{2} - 2}

w h a t i s m e a n t b y D_{f o g} ? (d e r i v a t i v e ?)

2 ⟩ 2 \int \sqrt{- [{(x - 1)}^{2} + 2 - 1]} d x

= 2 \int \sqrt{- {(x - 1)}^{2} - 1} d x

i t i s l i k e f i n d i n g \int \sqrt{- x} d x; I {h v n}^{'} t l e a r n t

a b o u t c o m p l e x i n t e g r a l s .

3 ⟩ f^{- 1} (x) = 3 - \frac{x^{2}}{2} = \frac{1}{2} (12 - x^{2})

I = \frac{1}{4} \int \frac{12 - x^{2}}{\sqrt{3 - x}} d x

\sqrt{3 - x} = t \Rightarrow x = 3 - t^{2} \Rightarrow d x = - 2 t d t

I = \frac{1}{4} \int \frac{12 - {(3 - t^{2})}^{2}}{t} (- 2 t) d t

= \frac{1}{2} \int [{(t^{2} - 3)}^{2} - 12] d t = \frac{1}{2} \int (t^{4} - 6 t^{2} - 3) d t

= \frac{1}{10} {(3 - x)}^{2} \sqrt{3 - x} - (3 - x) \sqrt{3 - x} - \frac{3}{2} \sqrt{3 - x} + C

= \frac{1}{10} (x^{2} - 5 x - 9) \sqrt{3 - x} + C

4 ⟩ c o m p l e x i n t e g r a l

Commented by mathmax by abdo last updated on 17/May/20

t h a n k y o u k u n a l

Commented by mathmax by abdo last updated on 17/May/20

D_{f o g} m e a n s e t o f d e f i n i t i o n . .!