FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.47Analytical Expression : Let α – h, α, α + h be the three values of x (h is very small) : then the corresponding
values of y will be f(α – h), f(α) and f(α + h). If however, f(α) be greater than f(α – h) and f(α + h), then f(x)
is said to be maximum at x = α. Again if f(α) be less than both f(α – h) and f(α + h), the f(x) is said to be
minimum at x = α.
Working Rule : by First Derivative Method :
Steps to find the maximum or minimum point of a curve y = f(x).
Find f ′ (x) and equate it to zero. From the equation f ′ (x) = 0, find the value of x, say α and β.
Here the number of roots of f ′ (x) = 0 will be equal to the number of degree of f ′ (x) = 0.
Then find f ′ (α – h) and f ′ (α + h), then note the change of sign if any (here h is very small).
If the change is from positive to negative, f(x) will be maximum at x = α. If again the change of sign is from
negative to positive, f(x) will be maximum at x = α.
Similar treatment for x = β.
Note : We have seen that dydx changes sign (positive to negative or vice versa) is passing through the value
zero. It may also happen that dydx changes sign in passing through an infinite value (the detail is not shown
as present).
Example 103: Examine for maximum and minimum for the function f(x) = x^3 – 27x + 10.
Solution:
Now f ′ (x) = 3x^2 – 27. For maximum and minimum f ′ (x) = 0 or 3x^2 – 27 = 0
or, x^2 =^237 =^9 ∴ x = ± 3.
Now let us enquire whether f(x) is maximum or minimum at these values of x.
For x = 3, let us assign to x, the values of 3 – h and 3 + h (h is very small) and put these values at f(x).
Now f ′ (3 – h) = 3 (3 – h)^2 – 27 which is negative for h is very small.
And f ′ (3 + h) = 3(3 + h)^2 – 27 which is positive.
Thus f ′ (x)
i.e.,dy
dx
changes sign from negative to positive as it passes throug x = 3. Therefore f(x) is minimum
at x = 3. The minimum value is
f(x) = 3^3 – 27.3 + 10 = 27 – 81 + 10 = – 44.
Similarly f ′ (– 3 – h) = 3 ( – 3 – h)^2 – 27, it is positive and f ′ ( – 3 + h) = 3 (– 3 +h )^2 – 27, it is negative, and
consequently the change of sign of f(x) being positive to negative, f(x) is maximum
at x = – 3.
The maximum value is f(– 3) = (– 3)^3 – 27 ( – 3) + 10 = – 27 + 81 + 10 = 64.
Working Rule :
By second Derivative Method :
For the function y = f(x), find dydx and make it zero. From the equation dydx= 0, find the values of x say a and
b.