 Mathematics First Derivative Test, Second Derivative Test And Maximum and Minimum Values of a Function in a Closed Interval For CBSE-NCERT

### Topic covered

star First Derivative Test
star Second Derivative Test
star Maximum and Minimum Values of a Function in a Closed Interval

### First Derivative Test \color{green} ✍️ Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I. Then

(i) If f ′(x) changes sign from positive to negative as x increases through c,
i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.

(ii) If f ′(x) changes sign from negative to positive as x increases through c,
i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.

(iii) If f ′(x) does not change sign as x increases through c,
then c is neither a point of local maxima nor a point of local minima.
Infact, such a point is called point of inflection.

"Note :" color{green}{" If c is a point of local maxima of f , then f (c) is a local maximum value of f. "}
color{green}{"Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f. "}
Q 3115156060 Find all points of local maxima and local minima of the function f
given by

f (x) = x^3 – 3x + 3 .
Class 12 Chapter 6 Example 29 Solution:

We have

f (x) = x^3 – 3x + 3

or f ′(x) = 3x^2 – 3 = 3 (x – 1) (x + 1)

or f ′(x) = 0 at x = 1 and x = – 1

Thus, x = ± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f . Let us first examine the point x = 1.
Note that for values close to 1 and to the right of 1, f ′(x) > 0 and for values close
to 1 and to the left of 1, f ′(x) < 0. Therefore, by first derivative test, x = 1 is a point
of local minima and local minimum value is f (1) = 1. In the case of x = –1, note that
f ′(x) > 0, for values close to and to the left of –1 and f ′(x) < 0, for values close to and
to the right of – 1. Therefore, by first derivative test, x = – 1 is a point of local maxima
and local maximum value is f (–1) = 5.
Q 3125156061 Find all the points of local maxima and local minima of the function f
given by

f (x) = 2x^3 – 6x^2 + 6x +5 .
Class 12 Chapter 6 Example 30 Solution:

We have

f (x) = 2x^3 – 6x^2 + 6x + 5

or f ′(x) = 6x^2 – 12x + 6 = 6 (x – 1)^2

or f ′(x) = 0 at x = 1

Thus, x = 1 is the only critical point of f . We shall now examine this point for local
maxima and/or local minima of f. Observe that f ′(x) ≥ 0, for all x ∈ R and in particular
f ′(x) > 0, for values close to 1 and to the left and to the right of 1. Therefore, by first
derivative test, the point x = 1 is neither a point of local maxima nor a point of local
minima. Hence x = 1 is a point of inflexion.

"Remark:" One may note that since f ′(x), in Example 30, never changes its sign on R, graph of f has no turning points and hence no point of local maxima or local minima.

### Second Derivative Test \color{green} ✍️ Let f be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c. Then

(i)color{orange}{x = c\ \ "is a point of local maxima if" \ \ f ′(c) = 0}
color{orange}{"and" f ″(c) < 0}
=> The value f (c) is local maximum value of f .

(ii)color{orange}{ x = c \ \ "is a point of local minima if" \ \ f ′(c) = 0}
color{orange}{"and" f ″(c) > 0}
=> In this case, f (c) is local minimum value of f .

(iii) color{orange}{"The test fails if" \ \ f ′(c) = 0 "and" f ″(c) = 0.}

"Note :" As f is twice differentiable at c, we mean second order derivative of f exists at c.
Q 3155156064 Find local minimum value of the function f
given by f (x) = 3 + |x |, x ∈ R.
Class 12 Chapter 6 Example 31 Solution:

Note that the given function is not differentiable
at x = 0. So, second derivative test fails. Let us try first
derivative test. Note that 0 is a critical point of f . Now
to the left of 0, f (x) = 3 – x and so f ′(x) = – 1 < 0. Also

to the right of 0, f (x) = 3 + x and so f ′(x) = 1 > 0. Therefore, by first derivative test,
x = 0 is a point of local minima of f and local minimum value of f is f (0) = 3.
Q 3175156066 Find local maximum and local minimum values of the function f given by

f (x) = 3x^4 + 4x^3 – 12x^2 + 12
Class 12 Chapter 6 Example 32 Solution:

We have

f (x) = 3x^4 + 4x^3 – 12x^2 + 12

or f ′(x) = 12x^3 + 12x^2 – 24x = 12x (x – 1) (x + 2)

or f ′(x) = 0 at x = 0, x = 1 and x = – 2.

Now f ″(x) = 36x^2 + 24x – 24 = 12 (3x^2 + 2x – 1)
or { tt ( (f'' (0) = -12 < 0), ( f'' (1) = 48 > 0 ), ( f '' (-2) = 84 > 0 ) )

Therefore, by second derivative test, x = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local
minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20,
respectively.
Q 3105156068 Find all the points of local maxima and local minima of the function f
given by

f (x) = 2x^3 – 6x^2 + 6x +5 .
Class 12 Chapter 6 Example 33 Solution:

We have

f (x) = 2x^3 – 6x^2 + 6x +5

or  { tt ( ( f' (x) = 6x^2 -12 x + 6 = 6 (x-1)^2 ), ( f'' (x) =12 (x-1 )) )

Now f ′(x) = 0 gives x =1. Also f ″(1) = 0. Therefore, the second derivative test
fails in this case. So, we shall go back to the first derivative test.
We have already seen (Example 30) that, using first derivative test, x =1 is neither
a point of local maxima nor a point of local minima and so it is a point of inflexion.
Q 3125256161 Find two positive numbers whose sum is 15 and the sum of whose
squares is minimum.
Class 12 Chapter 6 Example 34 Solution:

Let one of the numbers be x. Then the other number is (15 – x). Let S(x)
denote the sum of the squares of these numbers. Then

S(x) = x^2 + (15 – x)^2 = 2x^2 – 30x + 225

or { tt ( (S'(x) = 4x-30 ), (S'' (x) = 4) )

Now S′(x) = 0 gives x = 15/2 . Also S '' (15/2) = 4 >0 Therefore, by second derivative

text, x = 15/2 is the point of local minima of S. Hence the sum of squares of numbers is
minimum when the numbers are 15/2 and  15-15/2 = 15/2

"Remark:" Proceeding as in Example 34 one may prove that the two positive numbers,
whose sum is k and the sum of whose squares is minimum, are k/2 and k/2.
Q 3145256163 Find the shortest distance of the point (0, c) from the parabola y = x^2,
where 0 ≤ c ≤ 5.
Class 12 Chapter 6 Example 35 Solution:

Let (h, k) be any point on the parabola y = x2. Let D be the required distance
between (h, k) and (0, c). Then

D = sqrt ( (h-0)^2 + (k-c)^2 ) = sqrt (h^2 + (k-c)^2 ) ..........(1)

Since (h, k) lies on the parabola y = x^2, we have k = h^2. So (1) gives

D ≡ D(k) = sqrt ( k + (k-c)^2 )

or  D′(k) = (1+2 ( k-c) )/( 2 sqrt ( k + (k-c)^2 ) )

Now D′(k) = 0 gives  k = (2c -1)/2

Observe that when  k < (2c -1)/2, then  2(k − c) +1< 0 , i.e., D′(k) < 0 . Also when

k > (2c-1)/2 , then D' (k) > 0 . So, by first derivative test, D(k) is minimum at k = (2c -1)/2

Hence, the required shortest distance is given by

D ( (2c-1)/2) = sqrt ( (2c-1)/2 + ( (2c-1)/2 -c)^2 ) = (sqrt (4c -1) )/2

"Note :" The reader may note that in Example 35, we have used first derivative test instead of the second derivative test as the former is easy and short.
Q 3145456363 Let AP and BQ be two vertical poles at
points A and B, respectively. If AP = 16 m, BQ = 22 m
and AB = 20 m, then find the distance of a point R on
AB from the point A such that RP^2 + RQ^2 is minimum
Class 12 Chapter 6 Example 36 Solution:

Let R be a point on AB such that AR = x m.
Then RB = (20 – x) m (as AB = 20 m). From Fig 6.18,
we have

RP^2 = AR^2 + AP^2

and RQ^2 = RB^2 + BQ^2

Therefore RP^2 + RQ^2 = AR^2 + AP^2 + RB^2 + BQ^2

= x^2 + (16)^2 + (20 – x)^2 + (22)^2

= 2x^2 – 40x + 1140

Let S ≡ S(x) = RP^2 + RQ^2 = 2x^2 – 40x + 1140.

Therefore S′(x) = 4x – 40.

Now S′(x) = 0 gives x = 10. Also S″(x) = 4 > 0, for all x and so S″(10) > 0.
Therefore, by second derivative test, x = 10 is the point of local minima of S. Thus, the
distance of R from A on AB is AR = x =10 m.
Q 3145556463 If length of three sides of a trapezium other than base are equal to 10 cm,
then find the area of the trapezium when it is maximum.
Class 12 Chapter 6 Example 37 Solution:

The required trapezium is as given in Fig 6.19. Draw perpendiculars DP and

CQ on AB. Let AP = x cm. Note that ΔAPD ~ ΔBQC. Therefore, QB = x cm. Also, by

Pythagoras theorem, DP = QC = sqrt (100 - x^2)  Let A be the area of the trapezium. Then

A ≡ A(x) = 1/2  (sum of parallel sides) (height)

=1/2 (2x +10+10) (sqrt (100 - x^2) )

= (x+10) (sqrt (100 - x^2) )

or A' (x) = (x+10) (-2x)/( 2 sqrt (100 - x^2) ) + (sqrt (100 - x^2))

= (-2 x^2 -10x +100)/(sqrt (100 - x^2 ) )

Now A′(x) = 0 gives 2x^2 + 10x – 100 = 0, i.e., x = 5 and x = –10.
Since x represents distance, it can not be negative.
So, x = 5. Now

A '' (x) = (sqrt (100-x^2) (-4x -10) - (-2x^2 -10 x +100) (-2x)/( 2 sqrt (100-x^2) ) )/(100 - x^2 )

= (2 x^3 -300 x - 1000 )/( (100 - x^2)^3/2 ) (on simplification)

or A'' (5) = (2 (5)^3 -300 (5) - 1000 )/( (100- (5)^2 )^(3/2) ) = (-2250)/(75 sqrt 75) = (-30)/(sqrt 75) < 0

Thus, area of trapezium is maximum at x = 5 and the area is given by

A(5) = (5+10) sqrt (100 - (5)^2) = 15 sqrt (75) = 75 sqrt 3 cm^2
Q 3115556469 Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone.
Class 12 Chapter 6 Example 38 Solution:

Let OC = r be the radius of the cone and OA = h be its height. Let a cylinder
with radius OE = x inscribed in the given cone (Fig 6.20). The height QE of the cylinder
is given by

 (QE)/(OA) = (EC)/(OC)  (since ΔQEC ~ ΔAOC)

or  (QE)/h = (r-x)/r

or QE = ( h (r-x) )/r

Let S be the curved surface area of the given
cylinder. Then

S ≡ S(x) = (2 pi x h (r-x) )/r = (2 pi h)/r (rx - x^2)

or { tt ( (S'(x) = (2 pi h)/r (r- 2x) ), ( S'' (x) = (-4 pi h )/r) )

Now S′(x) = 0 gives x = r/2 Since S″(x) < 0 for all x, S'' (r/2) < 0 . So x = r/2 is a

point of maxima of S. Hence, the radius of the cylinder of greatest curved surface area
which can be inscribed in a given cone is half of that of the cone.

### Maximum and Minimum Values of a Function in a Closed Interval => Let us consider a function f given by f(x) = x + 2, x ∈ (0, 1)

=> we may note that the function even has neither a local maximum value nor a local minimum value in (0,1).

=> if we extend the domain of f to the closed interval [0, 1], then f still may not have a local maximum (minimum) values but it certainly does have maximum value 3 = f(1) and minimum value 2 = f(0).

=> The maximum value 3 of f at x = 1 is called "absolute maximum" value ("global maximum" or greatest value) of f on the interval
[0, 1].

=> Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum value (global minimum or least value) of f on [0, 1].

=> Consider the graph given in Fig, Observe that the function f has a local minima at x = b and local minimum value is f(b). The function also has a local maxima at x = c and local maximum value is f(c).

=> Also from the graph, it is evident that f has absolute maximum value f(a) and absolute minimum value f(d). and note that they different from local minimum and maximum value.

\color{green} ✍️ Theorem : Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

\color{green} ✍️ Theorem : Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then

(i) color{blue}{f ′(c) = 0} if f attains its absolute "maximum" value at c.
(ii) color{blue}{f ′(c) = 0} if f attains its absolute "minimum" value at c.

Working Rule :

Step 1: Find all critical points of f in the interval, i.e., find points x where either f' (x) = 0 or f is not differentiable.
Step 2: Take the end points of the interval.
Step 3: At all these points (listed in Step 1 and 2), calculate the values of f .
Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3. This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f .
Q 3125656561 Find the absolute maximum and minimum values of a function f given by

f (x) = 2x^3 – 15x^2 + 36x +1 on the interval [1, 5].
Class 12 Chapter 6 Example 39 Solution:

We have

f (x) = 2x^3 – 15x^2 + 36x + 1

or f ′(x) = 6x^2 – 30x + 36 = 6 (x – 3) (x – 2)

Note that f ′(x) = 0 gives x = 2 and x = 3.
We shall now evaluate the value of f at these points and at the end points of the

interval [1, 5], i.e., at x = 1, x = 2, x = 3 and at x = 5. So

f (1) = 2(1^3) – 15 (1^2) + 36 (1) + 1 = 24

f (2) = 2(2^3) – 15 (2^2) + 36 (2) + 1 = 29

f (3) = 2(3^3) – 15 (3^2) + 36 (3) + 1 = 28

f (5) = 2(5^3) – 15 (5^2) + 36 (5) + 1 = 56

Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1.
Q 3155656564 Find absolute maximum and minimum values of a function f given by

f(x) = 12 x^(4/3) - 6x^(1/3) , x ∈ [−1, 1]
Class 12 Chapter 6 Example 40 Solution:

We have

f(x) = 12 x^(4/3) - 6x^(1/3)

or f' (x) =16 x^(1/3) - 2/(x^(2/3) ) = (2 (8x-1) )/(x^(2/3))

Thus, f ′(x) = 0 gives x=1/8 Further note that f ′(x) is not defined at x = 0. So the
critical points are x = 0 and x =1/8 Now evaluating the value of f at critical points

x = 0 , 1/8 and at end points of the interval x = –1 and x = 1, we have

f(-1) = 12 (-1)^(4/3) -6 (-1)^(1/3) =18

f(0) = 12 (0) -6 (0) = 0
f(1/8)=12 (1/8)^(4/3) - 6 (1/8)^(1/3) = (-9)/4

f(1) = 12 (1)^(4/3) -6 (1)^(1/3) = 6

Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1

and absolute minimum value of f is (-9)/4 that occurs at x = 1/8
Q 3125856761 An Apache helicopter of enemy is flying along the curve given by
y = x^2 + 7. A soldier, placed at (3, 7), wants to shoot down the helicopter when it is
nearest to him. Find the nearest distance.
Class 12 Chapter 6 Example 41 Solution:

For each value of x, the helicopter’s position is at point (x, x^2 + 7).
Therefore, the distance between the helicopter and the soldier placed at (3,7) is

sqrt ( (x-3)^2 + (x^2 + 7 -7 )^2 ) , i.e. , sqrt ( (x-3)^2 + x^4 )

let f (x) = (x – 3)^2 + x^4

or f ′(x) = 2(x – 3) + 4x^3 = 2 (x – 1) (2x^2 + 2x + 3)

Thus, f ′(x) = 0 gives x = 1 or 2x^2 + 2x + 3 = 0 for which there are no real roots.
Also, there are no end points of the interval to be added to the set for which f ′ is zero,
i.e., there is only one point, namely, x = 1. The value of f at this point is given by
f (1) = (1 – 3)^2 + (1)^4 = 5. Thus, the distance between the solider and the helicopter is

sqrt (f(1)) = sqrt 5

Note that sqrt 5 is either a maximum value or a minimum value. Since

sqrt (f(0) ) = sqrt ( (0-3)^2 + (0)^4 ) = 3 > sqrt 5 ,

it follows that sqrt 5 is the minimum value of sqrt ( f(x) ) . Hence, sqrt 5 is the minimum
distance between the soldier and the helicopter. 