If f - R - R and g - R - R given by f x - x - and g x - x - then find fog-4-3 and gof -4-3-

Fog(x) =f(g(x))= [|x|] fog( 4/3) = [| 4/3|] fog( 4/3) =[4/3]= 1 gof(x) =g(f(x))= |[x]| gof( 4/3) = |[ 4/3]| gof( 4/3) = | 2| = 2 ...

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Byju's Answer

Standard XII

Mathematics

Monotonically Increasing Functions

If f : R → R ...

Question

If $f : R \to R$ and $g : R \to R$ given by $f (x) = [x]$ and $g (x) = | x |$ , then find $f o g (\frac{- 4}{3})$ and $g o f (\frac{- 4}{3})$ .

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Solution

$f o g (x) = f (g (x)) = [| x |] f o g (\frac{- 4}{3}) = [∣ ∣ \frac{- 4}{3} ∣ ∣] f o g (\frac{- 4}{3}) = [\frac{4}{3}] = 1$
$g o f (x) = g (f (x)) = | [x] | g o f (\frac{- 4}{3}) = ∣ ∣ [\frac{- 4}{3}] ∣ ∣ g o f (\frac{- 4}{3}) = | - 2 | = 2$

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Similar questions

Q. If the function

f : R \to R

be given by

f (x) = x^{2} + 2

and

g : R \to R

be given by

g (x) = \frac{x}{x - 1}, x \neq 1

, find fog and gof and hence find

f o g (2)

and

g o f (- 3)

If the function $f : R \to R$ be given by $f (x) = x^{2} + 2$ and $g : R \to R$ be given by $g (x) = \frac{x}{x - 1}, x \neq 1$ ,
find fog and gof and hence fing fog(2) and gof(-3).

Q. If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| –

x, \forall x \in R

. Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).

Q. Let f : R → R and g : R → R be defined by f(x) = x² and g(x) = x + 1. Show that fog ≠ gof.

Q. Let f : R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x − 1. Show that fog = gof = I_R.

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If f:R→R and g:R→R given by f(x)=[x] and g(x)=|x|, then find fog(−43) and gof(−43).

fog(x) =f(g(x))= [|x|]fog(−43) = [∣∣−43∣∣]fog(−43) =[43]= 1 gof(x) =g(f(x))= |[x]|gof(−43) = ∣∣[−43]∣∣gof(−43) = |−2| = 2

If $f : R \to R$ and $g : R \to R$ given by $f (x) = [x]$ and $g (x) = | x |$ , then find $f o g (\frac{- 4}{3})$ and $g o f (\frac{- 4}{3})$ .

$f o g (x) = f (g (x)) = [| x |] f o g (\frac{- 4}{3}) = [∣ ∣ \frac{- 4}{3} ∣ ∣] f o g (\frac{- 4}{3}) = [\frac{4}{3}] = 1$
$g o f (x) = g (f (x)) = | [x] | g o f (\frac{- 4}{3}) = ∣ ∣ [\frac{- 4}{3}] ∣ ∣ g o f (\frac{- 4}{3}) = | - 2 | = 2$