The very small change in a function with respect to one of its variable can be called as the derivative of a function. We know that by finding the slope of the curve at an arbitrary point will be the measure of derivative at that point. Consider the Distance-Time graph. In rate measure, the first derivative will help us to find the velocity at a given time where as the second derivative will help us to find the acceleration at a given time t. Apart from these when we draw the graph of the curve, we see that the curve increases or decreases as the value of x increases. The first derivative test help us to find the increasing or decreasing function of the curve in a particular interval. It also help us to find the point of maximum or minimum, from which we can find the local minima or local maxima. In this section let us see about the definition of first derivative test and few examples to understand the test.

## First Derivative Test for Local Extrema

First Derivative: For any function f (x) which is differentiable, the derivative f ' (x) is the slope of the tangent to the curve at any arbitrary point (x, y).

At a particular point (a, b) where b = f (a). The derivative f ' (a) will give us the slope of the tangent at the given point (a, b).

The slope of the tangent f ' (a) is positive if the tangent makes acute angle with the x-axis.

In the above graph Tangent AB makes acute angle with the x-axis, and hence the slope is positive. (i. e) f ' (x) > 0
The tangent CD makes obtuse angle with the x-axis, and hence the slope is negative. (i. e) f ' (x) < 0

First Derivative Test for Local Extrema: Let f be a differentiable function defined on an interval I and let a $\epsilon$ I, Then,
(a) x = a is a point of local maximum value of f, if
(i) f ' (a) = 0 and,
(ii) f ' (x) changes sign from positive to negative as x increases through a.
(i. e) f ' (x) > 0 at every point sufficiently close to and to the left of a and
f ' (x) < 0 at every point sufficiently close to and to the right of a.

(b) x = a is a point of local minimum value of f, if
(i) f ' (a) = 0 and,
(ii) f ' (x) changes sign from negative to positive as x increases through a.
(i. e) f ' (x) < 0 at every point sufficiently close to and to the left of a and
f ' (x) > 0 at every point sufficiently close to and to the right of a.

(c) If f ' (a) = 0 and f ' (x) does not change sign as x increases through a, that is f ' (x) has the same sign in the complete neighborhood of a, then a is neither a point of local maximum value nor a point of local minimum value. In fact, such a point is called a point of inflection.

### First Derivative Test for Local Extrema Algorithm

Step 1: Put y = f (x)

Step 2: Find $\frac{dy}{dx}$ = f ' (x)

Step 3: Equate $\frac{dy}{dx}$ = f ' (x) = 0 and solve the equation for real values of x.

The real values of x are called the critical points of the function f (x).

Let c1 , c2, c3, ..... be the critical points of the function f (x).

Step 4: If f ' (x) changes its sign from positive to negative as x increases through c1 , then the function attains a local maximum at x = c1 .

If f ' (x) changes its sign from negative to positive as x increases through c1 , then the function attains a local minimum at x = c1 .

If f ' (x) does not change sign as x increases throgh c1 , then x = c1 is neither a point of local maximum nor a point of local minimum.
In this case x = c1 is called the point of inflection.

Similarly we can deal with other values of x.

## First Derivative Test Examples

### Solved Examples

Question 1: Find all the points of local maxima and minima of the function
f ( x ) = x3 - 6 x2 + 9 x - 8
Solution:

Step 1 :              Let y = f (x) = x3 - 6 x2 + 9 x - 8

Step 2 : $\frac{dy}{dx}$ = f ' (x) = 3 x2 - 12 x + 9

Step 3 :                (f ' x) = 0
=> 3 x2 - 12 x + 9 = 0
=> 3 (x2 - 4 x + 3) = 0
=>        x2 - 4 x + 3 = 0
=>   (x - 3) (x - 1) = 0
=>                       x = 1 or x = 3

The critical points are x = 1 and x = 3.

Let us take a point to the left of x= 1 and to the right of x = 1.
when x = 0.9,  f ' ( 0.9 ) = 3 ( x - 3 ) ( x - 1 )
= 3 ( 0.9 - 3 ) ( 0.9 - 1 )
= 3 ( - ve) ( - ve )
= + ve  > 0
When x = 1.1, f ' ( 1.1 ) = 3 (  x - 3 ) ( x - 1 )
= 3 ( 1.1 - 3 ) ( 1.1 - 1 )
= 3 ( -ve ) ( +ve )
= -ve < 0

Therefore, f ' ( x ) changes its sign from positive to negative as x increases through 1.
Therefore, x = 1 is the point of local maximum value of f.

Let us take a point to the left of x= 3 and to the right of x = 3
when x = 2.9, f ' ( 2.9 ) = 3 (  x - 3 ) ( x - 1 )
= 3 ( 2.9 - 3 ) ( 2.9 -1 )
= 3 ( -ve ) ( +ve)
= -ve < 0
when x = 3.1, f ' ( 3.1 ) = 3 (  x - 3 ) ( x - 1 )
= 3 ( 3.1 - 3 ) ( 3.1 - 1 )
= 3 ( +ve ) ( +ve )
= +ve > 0
Therefore, f ' ( x ) changes its sign from negative to positive as x increases through 3.
Therefore, x = 3 is the point of local minimum value of f.

From the above graph we see that the function increases from left and reaches maximum at x= 1 and then decreases.
Later when it decreases after x = 1, reaches a minimum point when x = 3 and then increases.
therefore x = 1 is called the point of maximum. and x = 3 is the point of minimum.

Question 2: Find the points of local maxima and local minima, if any of the function
f (x ) = sin 2x - x, where - $\frac{\pi}{2}$ $\le$ x $\le$ $\frac{\pi}{2}$.
Also find the local maximum or local minimum values as the case may be.
Solution:

We have     y = f (x) = sin 2x - x
=>        f ' (x) = 2 cos 2x - 1
for local maximum or minimum, we have f ' (x) = 0
=> 2 cos 2x - 1 = 0
=>      2 cos 2x = 1
=>         cos 2x = $\frac{1}{2}$

=>              2 x = - $\frac{\pi}{3}$

or        2 x = $\frac{\pi}{3}$     [ since - $\frac{\pi}{2}$ $\le$ x $\le$ $\frac{\pi}{2}$., we get - $\pi$ $\le$ 2 x $\le$ $\pi$]

=>                 x = - $\frac{\pi}{6}$

or   x =  $\frac{\pi}{6}$ are the points of local minima or maxima.

Let us take a point to the left and to the right of x = - $\frac{\pi}{6}$

When x = - $\frac{\pi}{4}$,             we get f ' ( x ) = 2 cos 2x - 1

= 2 cos 2($\frac{\pi}{4}$) - 1

= 2 cos $\frac{\pi}{2}$ - 1

= 0 - 1
= -1 < 0
when x = 0, we get            f ' ( x ) = 2 cos 2x - 1
= 2 cos 0 - 1
= 2 - 1
= 1 > 0
therefore, f ' ( x ) changes its sign from negative to positive when it moves from the left of x = - $\frac{\pi}{6}$ to its right.
Therefore  x = - $\frac{\pi}{6}$ is the point of local minimum.
Local minimum value = f ( -$\frac{\pi}{6}$)

= sin 2(-$\frac{\pi}{6}$) - (-$\frac{\pi}{6}$)

= - sin 2($\frac{\pi}{6}$) + $\frac{\pi}{6}$

= - sin ($\frac{\pi}{3}$) +  $\frac{\pi}{6}$

= - $\frac{\sqrt{3}}{2}$$\frac{\pi}{6} Let us take a point to the left and to the right of x = \frac{\pi}{6} when x = 0, we get f ' ( x ) = 2 cos 2x - 1 = 2 cos 0 - 1 = 2 - 1 = 1 > 0 When x = \frac{\pi}{4} , we get f ' ( x ) = 2 cos 2x - 1 = 2 cos 2(\frac{\pi}{4}) - 1 = 2 cos \frac{\pi}{2} - 1 = 0 - 1 = -1 < 0 therefore, f ' ( x ) changes its sign from positive to negative when it moves from the left of x = \frac{\pi}{6} to its right. Therefore x = \frac{\pi}{6} is the point of local maximum. Local maximum value = f (\frac{\pi}{6}) = sin 2(\frac{\pi}{6}) - (\frac{\pi}{6}) = sin 2(\frac{\pi}{6}) - \frac{\pi}{6} = sin (\frac{\pi}{3}) - \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$\frac{\pi}{6}$