Let- fx-x- -2-x-1 -x-1- -1-x-2 and gx-x- -x-0sin x- 0-x- then the exhaustive domain of gfx is

F(x)= [x], 2≤x≤ 1 |x|+1, 1 lt;x≤2 So, nbsp; domain of nbsp;f(x)=[ 2,2 ] and range of nbsp;f(x)={ 2, 1 }∪ [1,3] Now, nbsp;g(x)= [x], π≤ ...

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Solving Linear Differential Equations of First Order

Let, fx=[x], ...

Question

Let, $f (x) = {\begin{matrix} [x], - 2 \leq x \leq - 1 | x | + 1, - 1 < x \leq 2 \end{matrix}$ and $g (x) = {\begin{matrix} [x], - π \leq x \leq 0 sin x, 0 < x \leq π \end{matrix},$ then the exhaustive domain of $g (f (x))$ is

[- 2, 0]

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[- 1, 2]

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[0, 2]

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k

f

f (x) =

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, x \neq \frac{π}{2} 3, x = \frac{π}{2} \end{matrix}

x =

\frac{π}{2}

λ

f (x) = {\begin{matrix} λ (x^{2} - 2 x), if x \leq 0 4 x + 1, if x > 0 \end{matrix}

x = 0

x = 1

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} [x], & - 2 \leq x \leq - \frac{1}{2} 2 x^{2} - 1, & - \frac{1}{2} < x \leq 2 \end{matrix};

[.]

f (x - 1)

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (1 + a x)^{\frac{1}{x}}, if x < 0 b, if x = 0 \frac{(x + c)^{\frac{1}{3}} - 1}{(x + 1)^{\frac{1}{2}} - 1}, if x > 0 \end{matrix}

x = 0

f (x) = {\begin{matrix} e^{- 1 / x^{2}}, x > 0 0, x \leq 0 \end{matrix}

f (x)

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