If f(x) = {(x2,when x<0),(x, when 0≤x≤1),(1/x, when x>1), find:\(

1.	If f(x) = {(x2,when x<0),(x, when 0≤x≤1),(1/x, when x>1), find:\( f(x) = \begin{cases} x^2,\text{when }x<0\\ x, \text{when } 0≤x≤1\\\frac{1}{x},\text {when} x>1 \end{cases} \), Find:i.f(\(\frac{1}{2}\))ii. f(–2) iii. f(1) iv. f(\(\sqrt 3\))v. f(\(\sqrt {-3}\))
Answer» Given, f(x) = {(x²,when x<0),(x, when 0≤x≤1),(1/x, when x>1) i. f(\(\frac{1}{2}\)) When 0 ≤ x ≤ 1, f(x) = x ∴ f(\(\frac{1}{2}\)) = \(\frac{1}{2}\) ii. f(–2) When x < 0, f(x) = x² ⇒ f(–2) = (–2)² ∴ f(–2) = 4 iii. f(1) When x ≥ 1, f(x) = \(\frac{1}{x}\) ⇒ f(1) = \(\frac{1}{1}\) ∴ f(1) = 1 iv. f(\(\sqrt 3\)) We have, \(\sqrt 3\) ≈ 1.732 > 1 When x ≥ 1, f(x) = \(\frac{1}{x}\) ∴ f(\(\sqrt 3\)) = \(\frac{1}{\sqrt 3}\) v. f(\(\sqrt {-3}\)) We know, \(\sqrt {-3}\) is not a real number and the function f(x) is defined only when x ∈ R. Thus, f(\(\sqrt {-3}\)) does not exist.

If f(x) = {(x2,when x<0),(x, when 0≤x≤1),(1/x, when x>1), find:\( f(x) = \begin{cases} x^2,\text{when }x<0\\ x, \text{when } 0≤x≤1\\\frac{1}{x},\text {when} x>1 \end{cases} \), Find:i.f(\(\frac{1}{2}\))ii. f(–2) iii. f(1) iv. f(\(\sqrt 3\))v. f(\(\sqrt {-3}\))

Answer»

Given,

f(x) = {(x²,when x<0),(x, when 0≤x≤1),(1/x, when x>1)

i. f(\(\frac{1}{2}\))

When 0 ≤ x ≤ 1,

f(x) = x

∴ f(\(\frac{1}{2}\)) = \(\frac{1}{2}\)

ii. f(–2)

When x < 0,

f(x) = x²

⇒ f(–2) = (–2)²

∴ f(–2) = 4

iii. f(1)

When x ≥ 1,

f(x) = \(\frac{1}{x}\)

⇒ f(1) = \(\frac{1}{1}\)

∴ f(1) = 1

iv. f(\(\sqrt 3\))

We have,

\(\sqrt 3\) ≈ 1.732 > 1

When x ≥ 1,

f(x) = \(\frac{1}{x}\)

∴ f(\(\sqrt 3\)) = \(\frac{1}{\sqrt 3}\)

v. f(\(\sqrt {-3}\))

We know,

\(\sqrt {-3}\) is not a real number and the function f(x) is defined only when x ∈ R.

Thus,

f(\(\sqrt {-3}\)) does not exist.