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How to Find the Inverse of a Function

57 let g(x)=1...

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57 let g(x)=1+x-[x] and f(x)={-1,x<0;0,x=0;1,x>0, then for all x, f(g(x)) is equal to

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g (x) = 1 + x - [x]

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1 x < 0 0 x = 0 1 x > 0 \end{matrix}

x, f {g (x)}

g (x) = 1 + x - [x]

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1 & x < 0 0, & x = 0 1, & x > 0 \end{matrix}

x

f [g (x)]

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1, x < 0 0, x = 0 1, x > 0 \end{matrix}

g (x) = sin x + cos x

, then points of discontinuity of

(f o g) (x)

(0, 2 π)

g (x) = 1 + x - [x]

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \begin{matrix} - 1 & x < 0 \end{matrix} \begin{matrix} 0 & x = 0 \end{matrix} \begin{matrix} 1 & x > 0 \end{matrix} \end{matrix}

x, f [g (x)]

Q. Let g(x) = 1 + x - [x] and f(x)=

⎧ ⎨ ⎩ \begin{matrix} - 1, & x < 0 0, & x = 0 1, & x > 0 \end{matrix}

, then for all x, f(g(x)) is equal to

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