Show that the function f(x) = |x| isa. strictly increasing on [0, ∞]

1.	Show that the function f(x) = \|x\| isa. strictly increasing on [0, ∞]b. strictly decreasing on [− ∞, 0]
Answer» For x > 0 Modulus will open with + sign f(x) = +x ⇒ f’(x) = +1 which is < 0 for x < 0 Modulus will open with -ve sign f’(x) = -x = > f’(x) = -1 which is > 0 hence f(x) is increasing in x > 0 and decreasing in x < 0

Show that the function f(x) = |x| isa. strictly increasing on [0, ∞]b. strictly decreasing on [− ∞, 0]

Answer»

For x > 0

Modulus will open with + sign

f(x) = +x

⇒ f’(x) = +1 which is < 0

for x < 0

Modulus will open with -ve sign

f’(x) = -x = > f’(x) = -1 which is > 0

hence f(x) is increasing in x > 0 and decreasing in x < 0