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Maxima and minima - Applications of Differentiation - Stationary Value of a function | 11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation

Chapter: 11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation

Stationary Value of a function

Let f(x) be a continuous function on [a, b] and differentiable in (a, b). f(x) is said to be stationary at x = a if f ' (a)=0.

Stationary Value of a function

Let f(x) be a continuous function on [a, b] and differentiable in (a, b). f(x) is said to be stationary at x = a if f ' (a)=0.

The stationary value of f(x) is f(a) . The point (a,f(a) ) is called stationary point.


In figure 6.9 the function y = f(x) has stationary at x = a,x = b and x = c.

At these points, dx/dy = 0 . The tangents at these points are parallel to x – axis.

NOTE

By drawing the graph of any function related to economics data, we can study the trend of the business related to the function and therefore, we can predict or forecast the business trend.

 

Example 6.18

Show that the function f(x) = x3 3x2  + 4x , x R is strictly increasing function on R.

Solution :

f(x) = x 3 3x 2  + 4x , x R

f'(x) = 3x2–6x+4

= 3x2–6x+3+1

= 3(x–1)2+1

> 0, for all xR

Therefore, the function f is strictly increasing on (-∞,∞).

 

Example 6.19

Find the interval in which the function f(x)=x2–4x+6 is strictly increasing and strictly decreasing.

Solution :

Given that f(x) =x2–4x+6

Differentiate with respect to x,

f' (x) = 2x–4

When f’  (x) = 0 2x–4=0 

x = 2.

Then the real line is divided into two intervals namely (–∞,2) and (2,∞)


[ To choose the sign of f (x) choose any values for x from the intervals ans substitute in f  ‘ (x) and get the sign.]


 

Example 6.20

Find the intervals in which the function f given by f(x)=4x3–6x2–72x+30 is increasing or decreasing.

Solution :

f(x) = 4x3–6x2–72x+30

f ‘(x) = 12x2–12x–72

= 12(x2x–6)

= 12(x–3)(x+2)

f (x) = 0 12(x–3)(x+2)=0

x= 3 (or) x = –2

f(x) has stationary at x =3 and at x = –2.

These points divides the whole interval into three intervals namely (–∞,–2),(–2,3) and (3,∞).


 

Example 6.21

Find the stationary value and the stationary points f(x)=x2+2x–5.

Solution :

Given that   f(x) = x2 + 2x – 5 … (1)

f'(x) = 2x + 2

At stationary points, f ‘(x) = 0

2x + 2 = 0

x = –1

f(x) has stationary value at x = –1

When x = –1, from (1)

f(–1) = (–1)2+2(–1)–5

= – 6

Stationary value of f (x) is – 6

Hence stationary point is (–1,–6)

 

Example 6.22

Find the stationary values and stationary points for the function

f(x)=2x3+9x2+12x+1.

Solution :

Given that f(x) = 2x3+9x2+12x+1.

f'(x) = 6x2+18x+12

= 6(x2+3x+2)

= 6(x+2)(x+1)

f'(x) = 0 6 (x+2)(x+1) = 0

x + 2 = 0 (or) x + 1 = 0.

x = –2 (or) x = –1

f(x) has stationary points at x = – 2 and x = – 1

Stationary values are obtained by putting x = – 2 and x = – 1

When x = – 2

f(–2) = 2(–8)+9(4)+12(–2)+1

= –3

When x = – 1

f(–1) = 2(–1)+9(1)+12(–1)+1

= –4

The stationary points are (–2,–3) and (–1,–4)

 

Example 6.23

The profit function of a firm in producing x units of a product is given by P(x)=  x3/3  + x2 + x. Check whether the firm is running a profitable business or not.

Solution :


It is clear that P'(x)>0 for all x.

The firm is running a profitable business.

 

IMPORTANT NOTE

Let R(x) and C(x) are revenue function and cost function respectively when x units of commodity is produced. If R(x) and C(x) are differentiable for all x > 0 then P(x) = R(x) – C(x) is maximized when Marginal Revenue = Marginal cost. That is, when Rl (x)= C (x) profit is maximum at its stationary point.

 

Example 6.24

Given C(x)= x2/6 +5x+200 and p(x) = 40–x are the cost price and selling price when x units of commodity are produced. Find the level of the production that maximize the profit.

Solution :



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11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation


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