If f ( x )= x and g ( x )=| x |, then ( f + g )( x ) is equal to

Question

If f ( x )= x and g ( x )=| x |, then ( f + g )( x ) is equal to (A) 0 for all x ∈ R (B) 2 x for all x ∈ R (C)  begincases 2x textfor x ge 0 [2ex] 0, textfor x < 0 endcases (D)  begincases 0, textfor x ge 0 [2ex] 2x, textfor x < 0 endcases

Tardigrade Admin · Accepted Answer

Given functions are: f ( x )= x and g ( x )=| x | ∴ (f+g)(x)=f(x)+g(x)=x+ mid x| According to definition of modulus function, (f + g) (x) = begincases x + x, x ge 0 [2ex] x - x, x < 0 endcases = begincases 2x, x ge 0 [2ex] 0, x < 0 endcases

Q. If $f (x) = x$ and $g (x) = ∣ x ∣$ , then $(f + g) (x)$ is equal to

$0$ for all $x \in R$

$2 x$ for all $x \in R$

$⎩ ⎨ ⎧ 2 x 0, for x \geq 0 for x < 0$

$⎩ ⎨ ⎧ 0, 2 x, for x \geq 0 for x < 0$

Given functions are: $f (x) = x$ and $g (x) = ∣ x ∣$
$∴ (f + g) (x) = f (x) + g (x) = x + ∣ x ∣$
According to definition of modulus function,
$(f + g) (x) = ⎩ ⎨ ⎧ x + x, x - x, x \geq 0 x < 0 = ⎩ ⎨ ⎧ 2 x, 0, x \geq 0 x < 0$

Q. If f(x)=x and g(x)=∣x∣, then (f+g)(x) is equal to

0 for all x∈R

2x for all x∈R

⎩⎨⎧​2x0,​forx≥0forx<0​

⎩⎨⎧​0,2x,​forx≥0forx<0​