Let f x be defined on - - 2-2- - and is given by f x - - - 1 - - 2 - x - 0- x - 1 - 0 - x - 2 - and g x - f - x - - - f x - - then find g x -

Now,for x⟩ 1 , f( x ) is always positive,So, | f( x )| = f( x ) So, [ | f( x )| = {[ 1, 2 ≤ x ≤ 0; 1 x,0⟨ x ≤ 1 ].; ...

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Byju's Answer

Standard IX

Mathematics

Elimination Method

Let f x be ...

Question

Let $f (x)$ be defined on $[- 2, 2]$ , and is given by $f (x) = {\begin{matrix} - 1, - 2 \leq x \leq 0 x - 1, 0 \leq x \leq 2 \end{matrix} a n d g (x) = f (| x |) + | f (x) |$ , then find $g (x)$ .

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Solution

now,for $x ⟩ 1, f (x)$ is always positive,
So, $| f (x) | = f (x)$
So, $\begin{matrix} | f (x) | = {\begin{matrix} 1, - 2 \leq x \leq 0 1 - x, 0 ⟨ x \leq 1 \end{matrix} x - 1, 1 \leq x \leq 2 \end{matrix}$
So, $\begin{matrix} g (x) = f (| x |) + | f (x) | = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 1 - 1 = 0, - 2 \leq x \leq 0 0, 0 ⟨ x \leq 1 2 x - 2, x ⟩ 1 \end{matrix} \end{matrix}$

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Let f(x) be defined on [−2,2] , and is given by f(x)={−1,−2≤x≤0x−1,0≤x≤2andg(x)=f(|x|)+|f(x)| , then find g(x).

now,for x⟩1,f(x)is always positive,So,|f(x)|=f(x)So,|f(x)|={1,−2≤x≤01−x,0⟨x≤1x−1,1≤x≤2So,g(x)=f(|x|)+|f(x)|=⎧⎪⎨⎪⎩1−1=0,−2≤x≤00,0⟨x≤12x−2,x⟩1

Let $f (x)$ be defined on $[- 2, 2]$ , and is given by $f (x) = {\begin{matrix} - 1, - 2 \leq x \leq 0 x - 1, 0 \leq x \leq 2 \end{matrix} a n d g (x) = f (| x |) + | f (x) |$ , then find $g (x)$ .