If f(x) =\(\begin{cases}x^2 & \quad \text{when } x < 0 \text{ }\\x& \

1.	If f(x) =\(\begin{cases}x^2 & \quad \text{when } x < 0 \text{ }\\x& \quad \text{when } 0 ≤ x < 1\text{}\\ 1/x, & \quad \text{when } x ≥ 1\text{}\end{cases}\)Find:(i) f (1/2)(ii) f (-2)(iii) f (1)(iv) f (√3)(v) f (√-3)
Answer» (i) f(1/2) If, 0 ≤ x ≤ 1, f(x) = x ∴ f (1/2) = 1/2 (ii) f(-2) If, x < 0, f(x) = x² f(–2) = (–2)² = 4 ∴ f (–2) = 4 (iii) f (1) If, x ≥ 1, f (x) = 1/x f(1) = 1/1 ∴ f(1) = 1 (iv) f (√3) Now, we have √3 = 1.732 > 1 If, x ≥ 1, f (x) = 1/x ∴ f (√3) = 1/√3 (v) f (√-3) As we know that √-3 is not a real number and the function f(x) is defined only when x ∈ R. Hence, f(√-3) does not exist.

If f(x) =\(\begin{cases}x^2 & \quad \text{when } x < 0 \text{ }\\x& \quad \text{when } 0 ≤ x < 1\text{}\\ 1/x, & \quad \text{when } x ≥ 1\text{}\end{cases}\)Find:(i) f (1/2)(ii) f (-2)(iii) f (1)(iv) f (√3)(v) f (√-3)

Answer»

(i) f(1/2)

If, 0 ≤ x ≤ 1, f(x) = x

∴ f (1/2) = 1/2

(ii) f(-2)

If, x < 0, f(x) = x²

f(–2) = (–2)²

= 4

∴ f (–2) = 4

(iii) f (1)

If, x ≥ 1, f (x) = 1/x

f(1) = 1/1

∴ f(1) = 1

(iv) f (√3)

Now, we have √3 = 1.732 > 1

If, x ≥ 1, f (x) = 1/x

∴ f (√3) = 1/√3

(v) f (√-3)

As we know that √-3 is not a real number and the function f(x) is defined only when x ∈ R.

Hence, f(√-3) does not exist.