Let $g(x) = 1 + x - \lbrack x brack$ and $f(x) = \left\{ \begin{array}{r} \begin{array}{r} - 1,\ \ \ \ \ \ \ \ \ \ x 0\ \ \ \end{array} ight.\ then\ for\$ for $x,f\lbrack g(x) brack$ is equal to

Question

Let $g(x) = 1 + x - \lbrack xbrack$ and $f(x) = \left\{ \begin{array}{r} \begin{array}{r} - 1,\ \ \ \ \ \ \ \ \ \ x < 0\ \ \ \ \ \ \ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ x = 0,\ \end{array} \ 1\ \ \ \ \ \ \ \ \ \ \ \ x > 0\ \ \ \end{array} ight.\ then\ for\$ for $x,f\lbrack g(x)brack$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$g(x) = 1 + x - \lbrack x\rbrack$ is greater than 1 since $x - \lbrack x\rbrack > 0,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f\left\{ g(x) \right\} = 1$

Q. Let $g(x) = 1 + x - \lbrack x\rbrack$ and $f(x) = \left\{ \begin{array}{r} \begin{array}{r} - 1,\ \ \ \ \ \ \ \ \ \ x < 0\ \ \ \ \ \ \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ x = 0,\ \end{array} \\ 1\ \ \ \ \ \ \ \ \ \ \ \ x > 0\ \ \ \end{array} \right.\ then\ for\$ for $x,f\lbrack g(x)\rbrack$ is equal to

Explanation

Reviewed by Dr. Abubacker

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