If f(x) = {(x^2, when x < 0), (x, when 0 ≤ x < 1), (1/x, when x ≥ 1) Find: (i) f (1/2)

Question

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 1

(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.