Let y = f (x) be a function defined at x = a and also in the vicinity of the point x = a. Then, f (x) is said to have a local maximum at x = a , if the value of the function at x = a is greater than the value of the function at the neighbouring points of x = a .

Mathematically, f (a) > f (a – h) and f (a) > f (a + h) where h > 0.

Similarly, 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 neighbouring points of x = a.

Mathematically, f(a) < f(a – h) and f(a) < f (a + h) where h > 0.

**A local maximum or a local minimum is also called a local extremum.**

__Test for Local Maximum / Minimum:__

We have two cases to consider:

Test for Local Maximum/Minimum at x = a if f (x) is Differentiable at x = a :

If f (x) is differentiable at x = a and if it is the critical point of the function (i.e. f ‘(a) = 0) then we have the following three tests to decide whether f(x) has a local maximum or local minimum or neither at x = a .

__First Derivative Test:__

If f'(a) = 0 and f ‘(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 ‘(a – 0) > 0 and f ‘(a + 0) < 0. It means that f ‘(x) should change it’s sign from positive to negative.

(ii) f(x) would have local minimum at x = a if f ‘(a – 0) < 0 and f ‘(a+0) > 0. It means that f ‘(x) should change it’s sign from negative to positive.

(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.

__Second Derivative Test:__

This test is basically the mathematical representation of the first derivative test. It simply says that,

(i) If f ‘(a) = 0 and f ”(a) < 0 , then f (x) would have a local maximum at x = a.

(ii) If f ‘(a) = 0 and f ”(a) > 0 , then f (x) would have a local minimum at x = a.

(iii) If f ‘(a) = 0 and f ”(a) = 0, then this test fails and the existence of a local maximum/minimum at x = a is decided on the basis of the nth derivative test.

__nth Derivative Test:__

It is nothing but the general version of the second derivative test. It says that if, f ‘(a) = f ”(a) = f ”'(a) …. = f ^{n}(a) = 0 and f ^{n+1}(a) ≠ 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 f ^{n+1} (a) < 0 and would be a point of local minima if f ^{n+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 becomes very difficult to decide whether f'(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.

**Test for Local Maximum/Minimum at x = a if f (x) is not Differentiable at x = a:**

**Case (i) :** When f(x) is continuous at x = a and f ‘(a – h) and f'(a + h) exist and are non-zero, then f(x) has a local maximum or minimum at x = a if f ‘a – h) and f ‘(a + h) are of opposite signs.

If f ‘(a – h) > 0 and (a + h) < 0 then x = a will be a point of local maximum.

If f ‘(a – h) < 0 and (a + h) > 0 then x = a will be a point of local minimum.

**Case (ii) :** When f(x) is continuous and f ‘(a – h) and f ‘(a + h) exist but one of them is zero, then x = a is point of neither maxima nor minima.

**Case (iii) :** If f(x) is not continuous at x = a and f ‘ (a – h) and/or f ‘(a + h) are not finite, then compare the values of f(x) at the neighbouring points of x = a.

Remark:

∎ It is advisable to draw the graph of the function in the vicinity of the point x = a because the graph would give us the clear picture about the existence of local maxima/minima at x = a.

Solved Example : $ \large If \; f(x) = \left\{\begin{array}{ll} x ^2 \; , x \leq 0 \\ 2sinx \; , x > 2 \end{array} \right. $ , investigate the function at x = 0 for maxima/minima.

Solution: Analysing the graph of f(x) we get x = 0 is a point of minima.

**Alternately:**

$ \large f'(x) = \left\{\begin{array}{lll} 2x \; , x < 0 \\ non-diff \; , x = 0 \\ 2 cosx \; , x > 0 \end{array} \right. $

So x = 0 is a critical point f(0^{–}) > 0 as well as f(0^{+}) > 0 and f(0) = 0 hence a point of local minima.

**Note: **We cannot say that change of sign of derivative helps to determine minima/ maxima if f (x) is not differentiable at x = a.

Example: Let $ \large f(x) = \left\{\begin{array}{lll} x^2 \; , x < 0 \\ 5 \; , x = 0 \\ 2 sinx \; , x > 0 \end{array} \right. $

$ \large f ‘(x) = \left\{\begin{array}{lll} 2x \; , x < 0 \\ non-diff \; , x = 0 \\ 2 cosx \; , x > 0 \end{array} \right. $

Here also the derivative is changing sign in same manner but the point x = 0 is a point of maxima as f(0^{–}) < f(0) and f(0^{+}) < f(0).

Illustration : Find the coordinate of the point on y^{2} = 8x which is closest from x^{2} + (y + 6)^{2} = 1

Solution: Let Point on parabola y^{2} = 8x be (2α^{2}, 4α), centre of circle = (0, -6)

Distance between centre of circle and point on parabola

$\large S = \sqrt{(2\alpha^2)^2 + (4\alpha + 6)^2}$

D = 4α^{4} +(4α +6)^{2} where D = s^{2}

=>$\frac{dD}{d\alpha} =16 \alpha^3 + 2(4\alpha + 6) \times 4 = 0 $

2 α^{3} + 4α +6 = 0

=> α^{3} + 2α +3 = 0

α^{3} + α^{2}– α^{2} – α + 3α + 3 = 0

=> (α^{2}-α +3) (α + 1) = 0

=> α = -1

Now ,$\frac{d^2D}{d\alpha^2} = 48 \alpha^2 + 32 > 0 $ ; at α = -1

=> D is minimum at α = -1

Point on parabola (2, -4)

Solved Example : Let $ \large f(x) = \left\{\begin{array}{ll} x^3 + x ^2 + 10x \; , x < 0 \\ -3sinx \; , x \ge 0 \end{array} \right. $

Investigate x = 0 for local maxima/minima.

Solution: Clearly f (x) is continuous at x = 0 but not differentiable at x = 0 as f(0) = f (0 –0) = f (0 + 0) = 0

$f_{-}'(0) = lim_{h \rightarrow 0} \frac{f(-h) – f(0)}{-h} $

$ = lim_{h \rightarrow 0} \frac{-h^3 + h^2 -10 h – 0}{-h} $

= 10

But , $f_{+}'(0) = lim_{h \rightarrow 0} \frac{f(h) – f(0)}{h} $

$ = lim_{h \rightarrow 0} \frac{- 3 sin h}{h}$

= – 3

Since $f_{-}'(0) > 0 $ and $f_{+}'(0) < 0 $ , x = 0 is the point of local maximum.

Solved Example : Let $ \large f(x) = \left\{\begin{array}{ll} 3 \; , x \le 1 \\ 4 -x \; , x > 1 \end{array} \right. $

Investigate x = 1 for the existence of a local maximum / minimum

**Solution :** $f_{-}'(1) = lim_{h \rightarrow 0} \frac{3 – 3}{-h} = 0 $

$f_{+}'(1) = lim_{h \rightarrow 0} \frac{4 -(1+h) – 3}{h} = -1 $

It shows that x = 1 is neither a point of maximum nor minimum .