Paper 4: Fundamentals of Business Mathematics & Statistic

3.4 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

Example15 :

If y = x^2 + 4 is a function under consideration then solving for y = 0, we get

0 = x^2 + 4

or, x^2 = – 4

or, x= ± − 4

or, x 2 1 2i= ± − = ±

The number ±2i is a complex number whose real part is 0 & imaginary part is ±2

Example16 : Given f(x) = 2x^2 – 3x + 1 ; find f(2), f(0), f (– 3)

f(2) = 2.2^2 – 3.2 + 1 = 2.4 – 6 + 1 = 8 – 6 + 1 = 3

f(0) = 2.0^2 – 3.0 + 1 = 2.0 – 0 + 1 = 0 – 0 + 1 = 1

f( – 3) = 2 (– 3)^2 – 3. (– 3) + 1 = 2.9 + 9 + 1 = 18 + 9 + 1 = 28.

Example17 : If y = 4x – 1, find the value of y for x = 2. Can y be regarded as a function of x? Also find the

domain.

For x = 2, y = 4.2 – 1 = 8 – 1 = 7. Again for x = 0, y = – 1 and for x = – 1, y = – 5. So for every value of x in – ∞ <

x < ∞, we find different values of y, So y is a function of x and its domain is -

• ∞ < x < + ∞.
  Example18 : If f (x) = x + | x |, find f (3) and f (– 3) and show also they are not equal.
  f(3) = 3 + | 3 | = 3 + 3 = 6 ; f ( – 3) = – 3 + | – 3| = – 3 + 3 = 0.
  As 6 ≠ 0, so f(3) ≠ f ( – 3).
  Note : If f(x) = f(– x) [i.e., f (3) = f (– 3)] then f(x) will be an even function of x.
  Example19 : Show that x 5x 4^2 − + is not defined for 1 < x < 4

x 5x 4^2 − +^ = (x 1 x 4 .−)( − )

Now for any value x > 1, but < 4 the expression becomes imaginary. So the expression is undefined for

1 < x < 4.

Example20 : Find the domain of f(x)=x 9 2 x−

Here f(x) has a unique value except for x = 3, – 3.

For f 3( )=9 9 0−3 3= (undefined) and f 3(− )=9 9 0−−3 3=− (undefined)

∴ domain of the function f(x) is – ∞ < x < – 3 ; – 3 < x < 3 and 3 < x < ∞.

Example21 : Given the function

f x 5 1,( )= −2x− – 1 ≤ x < 0

x 2^2

= x 2−− , 0 ≤ x < 1