Mathematics First Derivative Test, Second Derivative Test And Maximum and Minimum Values of a Function in a Closed Interval For CBSE-NCERT
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Class 12 Chapter 6 - APPLICATION OF DERIVATIVES

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

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`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.
 
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