Maxima & Minima

LOCAL MAXIMUM

A function f(x) is said to have a local maximum at x=a if the value of f(a) is greater than all the values of f(x) in a small neighbourhood of x=a. Mathematically, f (a) > f (a – h) and f (a) > f (a + h) where h > 0, then a is called the point of local maximum.

LOCAL MAXIMUM

A function f(x) is said to have a local minimum at x = a, if the value of the function at x = a is less than the value of the function at the neighboring points of x = a. Mathematically, f (a) < f (a – h) and f (a) < f (a + h) where h > 0, then a is called the point of local minimum. A point of local maximum or a local minimum is also called a point of local extremum.

WORKING RULE TO DETERMINE THE POINTS OF LOCAL MAXIMA AND LOCAL MINIMA

1. First Derivative Test: If $f^{\prime }(a)$= 0 and $f^{\prime }(x)$changes it's sign while passing through the point x = a , then

(i) f(x) would have a local maximum at x = a if $f^{\prime }(a-0)>0$and $f^{\prime }(a+0)<0$. It means that $f^{\prime }(x)$should change it's sign from positive to negative. e.g. f (x) = –x² has local maxima at x = 0.

(ii) f(x) would have local minimum at x = a if $f^{\prime }(a-0)<0$and $f^{\prime }(a+0)>0$. It means that $f^{\prime }(x)$should change it's sign from negative to positive. e.g. f (x) = x² has local minima at x = 0.

(iii) If f(x) doesn't change it's sign while passing through x = a, then f (x) would have neither a maximum nor minimum at x = a. e.g. f (x) = x³ doesn't have any local maxima or minima at x = 0.

2. Second Derivative Test:

Step I: Let f(x) be a differentiable function on a given interval and let f'' be continuous at stationary point. Find f ' (x) and solve the equation f ' (x) = 0 given let x = a, b, … be solutions.

Step II: Case (i) : If f '' (a) < 0 then f(a) is maximum. (Refer to above figure)

Case (ii): If f ''(a) > 0 then f(a) is minimum. (Refer the above figure)

Note:

(i) If f''(a) = 0 the second derivatives test fails in that case we have to go back to the first derivative test.

(ii) If f''(a) = 0 and a is not a point of local maximum nor local minimum then a is a point of inflection.

3. n^th Derivative Test: It is nothing but the general version of the second derivative test. It says that if,

f' (a) = f''(a) = f'''(a) =………. = fⁿ (a) = 0 and fⁿ⁺¹ (a) $\ne$ 0 (all derivatives of the function up to order 'n' vanishes and (n + 1)th order derivative does not vanish at x = a), then f (x) would have a local maximum or minimum at x = a iff n is odd natural number and that x = a would be a point of local maxima if fⁿ⁺¹ (a) < 0 and would be a point of local minima if fⁿ⁺¹ (a) > 0.

It is clear that the last two tests are basically the mathematical representation of the first derivative test. But that shouldn't diminish the importance of these tests. Because at time it's becomes very difficult to decide whether $f^{\prime }(x)$ changes it's sign or not while passing through point x = a, and the remaining tests may come handy in these type of situations.

Illustration 1: Find the local maximum and local minimum for the following function :

(i) $f\left(x\right)=2x^3-21x^2+36x-20$

(ii) $f\left(x\right)=-x+2\sin x,0\le x\le 2\pi$

Solution : (i) $f\left(x\right)=2x^3-21x^2+36x-20$

$\therefore f^{\prime }\left(x\right)=3x^2-12x+9$

For local max. and local min. $f^{\prime }\left(x\right)=0\text{;k }3x^2-12x+9=0$

or $3\left(x^2-4x+3\right)=0or{x}^2-4x+3=0$

or $\left(x-1\right)\left(x-3\right)=0,\therefore x=1,x=3$

$f^{\prime \prime }\left(x\right)=6x-12$

at $x=1,f^{\prime \prime }\left(x\right)=6\times 1-12=-6<0$negative

at $x=3,f^{\prime \prime }\left(x\right)=6\times 3-12=6>0$positive

$\therefore atx=1,f\left(x\right)$ has maximum,

$f\left(1\right)=1-6+9+7=11$

$atx=3,f\left(x\right)$ has minimum,

$f\left(3\right)=27-54+27+7=7$

(ii) $f\left(x\right)=2x^3-21x^2+36x-20$

$f^{\prime }\left(x\right)=6x^2-42x+36=6\left(x^2-7x+6\right)$

$=6\left(x-1\right)\left(x-6\right)$

For local max. and local min.

$f^{\prime }\left(x\right)=0\Rightarrow 6\left(x-1\right)\left(x-6\right)=0$

$\therefore x=1,6$

$f^{\prime \prime }\left(x\right)=12x-42=6\left(2x-7\right)$

$\therefore f^{\prime \prime }\left(1\right)=6\left(2-7\right)=-30<0$

$f^{\prime \prime }\left(6\right)=6\left(12-7\right)=30>0$

$\therefore$ at x = 1, f (x) has local max and at x = 6 there is local min,

Local max value $=f\left(1\right)=2{\left(1\right)}^3-21{\left(1\right)}^2+36\left(1\right)-20$

$=2-21+36-20=-3$

Local min value $=f\left(6\right)=2\left(216\right)-21\left(36\right)+36\left(6\right)-20$

$=432-756+216-20=-128$

(iii) $f\left(x\right)=-x+2\sin x$

$\therefore f^{\prime }\left(x\right)=-1+2\cos x$

For local max. and local min. $f^{\prime }\left(x\right)=0$

$\therefore -1+2\cos x=0or\cos x=\frac{1}{2}$

$=\cos \frac{\pi }{3}=\cos \left(2\pi -\frac{\pi }{3}\right)=\cos \frac{5\pi }{3}$

$\therefore x=\frac{\pi }{3},\frac{5\pi }{3}$

$f^{\prime \prime }\left(x\right)=-2\sin x$

$\therefore f^{\prime \prime }\left(\frac{\pi }{3}\right)=-2\sin \frac{\pi }{3}=-2\left(\frac{\sqrt{3}}{2}\right)=-\sqrt{3}<0$

$\therefore f^{\prime \prime }\left(\frac{5\pi }{3}\right)=-2\sin \frac{5\pi }{3}=-2\left(\frac{-\sqrt{3}}{2}\right)=\sqrt{3}>0$

So at $x=\frac{\pi }{3}$the function has local max and its value is

$=f\left(\frac{\pi }{3}\right)=-\frac{\pi }{3}+2\sin \frac{\pi }{3}$

$=-\frac{\pi }{3}+2\times \frac{\sqrt{3}}{2}=-\frac{\pi }{3}+\sqrt{3}$

So at $x=\frac{5\pi }{3}$the function has local min and its value is

$=f\left(\frac{5\pi }{3}\right)=-\frac{5\pi }{3}+2\sin \frac{5\pi }{3}$

$=-\frac{5\pi }{3}-2\sin \frac{\pi }{3}=-\frac{5\pi }{3}-2\times \frac{\sqrt{3}}{2}$$=-\frac{5\pi }{3}-\sqrt{3}$

Global maxima or minima in [a, b]

Global maxima or minima of f(x) in [a, b] is basically the greatest or least value of f(x) in [a, b].

Global maxima or minima in [a, b] will always occur either at the critical points of f(x) within [a, b] or at the end points of the interval.

Step to find out the global maxima or minima in [a, b]

Step 1: Find out all the critical points of f(x) in (a, b). Let C₁, C₂,….C_n be the different critical points.

Step 2: Find the value of the function at these critical points and also at the end points of the domain. Let the values of the function at critical points be f(C₁), f(C₂)………..f(C_n).

Step 3: Find M₁ =max{ f(a), f(C₁), f(C₂)………..f(C_n), f(b)} and M₂= min{ f(a), f(C₁), f(C₂)………..f(C_n), f(b)}. Now M₁is the maximum value of f(x) in [a, b] and M₂ is the minimum value of f(x) in [a, b].

Global maxima or minima in (a, b):

To find the global maxima and minima in (a, b) step 1 and 2 is same but after that we have to be little bit careful. After step 1 and 2 find M₁ =max{ f(C₁), f(C₂)………..f(C_n)} and M₂= min{f(C₁), f(C₂)………..f(C_n)}.

Now if $\underset{\begin{array}{c}x\to a^{+}\\ orx\to b^{-}\end{array}}{\lim }f(x)>M_1$ $or\underset{\begin{array}{c}x\to a^{+}\\ orx\to b^{-}\end{array}}{\lim f(x)}<M_2$, f(x) would not have global maximum(or global minimum) in (a, b) but if $\underset{\begin{array}{c}x\to a^{+}\\ andx\to b^{-}\end{array}}{\lim }f(x)<M_1$ and $\underset{\begin{array}{c}x\to a^{+}\\ andx\to b^{-}\end{array}}{\lim f(x)}>M_2$ then M1 and M2 would respectively be the global maximum and global minimum of f(x) in (a,b)

Illustration 2: Let f (x)= 2x³ – 9x² + 12x + 6. Discuss the global maxima and minima of f (x) in [0, 2] and (1, 3).

Solution: f (x) = 2x³ – 9x² + 12 x + 6

$\Rightarrow f^{\prime }$(x) = 6x² – 18x + 12 = 6 (x² – 3x + 2) = 6 (x-1) (x-2)

First of all let us discuss [0, 2].

Clearly the critical point of f (x) in [0, 2] is x = 1.

f (0) = 6, f (1) = 11, f (2) = 10

Thus x = 0 is the point of global minimum of f(x) in [0, 2] and x = 1 is the point of global maximum.

Now let us consider (1, 3).

Clearly x = 2 is the only critical point in (1, 3).

f (2) = 10. $\underset{x\to 1+0}{\mathrm{Lim}}$f (x) = 11 and $\underset{x\to 3-0}{\mathrm{Lim}}$f (x) = 15.

Thus x = 2 is the point of global minimum in (1, 3) and the global maximum in

(1, 3) does not exist.