5. Algebra of real functions

Algebra of real functions

Examples:

Q.1: Write the range of a Signum function.

Solution:

The real function f: R → R defined by

is called the signum function. Domain of f = R, Range of f = {1, 0, – 1}

Q.2: Express the function f: A—R. f(x) = x² – 1. where A = { -4, 0, 1, 4) as a set of ordered pairs.

Solution:

Given,

A = {-4, 0, 1, 4}

f(x) = x² – 1

f(-4) = (-4)² – 1 = 16 – 1=15

f(0) = (0)² – 1 = -1

f(1) = (1)² – 1 = 0

f(4) = (4)² – 1 = 16 – 1 =15

Therefore, the set of ordered pairs = {(-4, 15), (0, -1), (1, 0), (4, 15)}

Q.3: Let f(x) = x² and g(x) = 2x + 1 be two real functions. Find

(f + g) (x), (f –g) (x), (fg) (x), (f/g ) (x)

Solution:

Given,

f(x) = x² and g(x) = 2x + 1

(f + g) (x) = x² + 2x + 1

(f – g) (x) = x² -(2x + 1) = x² – 2x – 1

(fg) (x) = x²(2x + 1) = 2x³ + x²

(f/g) (x) = x²/(2x + 1), x ≠ -1/2

Q.4: Redefine the function: f(x) = |x – 1| – |x + 6|. Write its domain also.

Solution:

Given function is f(x) = |x – 1| – |x + 6|

Redefine of the function is:

The domain of this function is R.

Q.5: The function f is defined by

Draw the graph of f(x).

Solution:

f(x) = 1 – x, x < 0, this gives

f(– 4) = 1 – (– 4)= 5;

f(– 3) =1 – (– 3) = 4,

f(– 2) = 1 – (– 2)= 3

f(–1) = 1 – (–1) = 2; etc,

Also, f(1) = 2, f (2) = 3, f (3) = 4, f(4) = 5 and so on for f(x) = x + 1, x > 0.

Thus, the graph of f is as shown in the below figure.

Q.6: Find the domain and range of the real function f(x) = x/1+x².

Solution:

Given real function is f(x) = x/1+x².

1 + x² ≠ 0

x² ≠ -1

Domain : x ∈ R

Let f(x) = y

y = x/1+x²

⇒ x = y(1 + x²)

⇒ yx² – x + y = 0

This is quadratic equation with real roots.

(-1)² – 4(y)(y) ≥ 0

1 – 4y² ≥ 0

⇒ 4y² ≤ 1

⇒ y² ≤1/4

⇒ -½ ≤ y ≤ ½

⇒ -1/2 ≤ f(x) ≤ ½

Range = [-½, ½]