The function f (x) = max(1- x), (1+ x), 2 is equivalent to

Question

The function f (x) = max(1- x), (1+ x), 2 is equivalent to (A) f(x) = begincases 1 -x x le - 1 2 , -1 < x < 1 1 + x, x ge - 1 endcases  (B) f(x) = begincases 1 + x x le - 1 2 , -1 < x < 1 1 - x, x ge 1 endcases  (C) f(x) = begincases 1 -x x le - 1 2 , -1 < x < 1 1 + x, x ge 1 endcases  (D) none of these

Tardigrade Admin · Accepted Answer

We draw the graphs of y = 1 - x, y = 1 + x and y = 2 From the graph we get f(x) = begincases 1 -x textif x le - 1 2 textif -1 < x < 1 1 +x textif x ge 1 endcases

Q. The function f (x) = max{(1- x), (1+ x), 2} is equivalent to

$f (x) = ⎩ ⎨ ⎧ 1 - x 2, 1 + x, x \leq - 1 - 1 < x < 1 x \geq - 1$

$f (x) = ⎩ ⎨ ⎧ 1 + x 2, 1 - x, x \leq - 1 - 1 < x < 1 x \geq 1$

$f (x) = ⎩ ⎨ ⎧ 1 - x 2, 1 + x, x \leq - 1 - 1 < x < 1 x \geq 1$

none of these

We draw the graphs of y = 1 - x, y = 1 + x and y = 2
From the graph we get
$f (x) = ⎩ ⎨ ⎧ 1 - x 2 1 + x if x \leq - 1 if - 1 < x < 1 if x \geq 1$

Q. The function f (x) = max{(1- x), (1+ x), 2} is equivalent to

f(x)=⎩⎨⎧​1−x2,1+x,​x≤−1−1<x<1x≥−1​

f(x)=⎩⎨⎧​1+x2,1−x,​x≤−1−1<x<1x≥1​

f(x)=⎩⎨⎧​1−x2,1+x,​x≤−1−1<x<1x≥1​

none of these

We draw the graphs of y = 1 - x, y = 1 + x and y = 2 From the graph we get f(x)=⎩⎨⎧​1−x2 1+x​if x≤−1if −1<x<1if x≥1​

$f (x) = ⎩ ⎨ ⎧ 1 - x 2, 1 + x, x \leq - 1 - 1 < x < 1 x \geq - 1$

$f (x) = ⎩ ⎨ ⎧ 1 + x 2, 1 - x, x \leq - 1 - 1 < x < 1 x \geq 1$

$f (x) = ⎩ ⎨ ⎧ 1 - x 2, 1 + x, x \leq - 1 - 1 < x < 1 x \geq 1$

We draw the graphs of y = 1 - x, y = 1 + x and y = 2
From the graph we get
$f (x) = ⎩ ⎨ ⎧ 1 - x 2 1 + x if x \leq - 1 if - 1 < x < 1 if x \geq 1$